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Stat 216 Exam Two Review Problems

1.
A random sample of 30 cigarettes of a certain brand has an average nicotine content of 2.6 milligrams and a standard deviation of 0.9 mg.
(a)
Give a point estimate of the true mean.
(b)
Construct a 99% confidence interval for the true average nicotine content of this particular brand of cigarettes. [Go to answer]

2.
The owner of a gasoline service station wants to study purchasing habits of motorists at his station. The owner speculates that motorists buy more than 11 gallons of gas, on the average. A random sample of 60 motorists during a certain week is selected with the following results: an average of 11.3 gallons purchased with a standard deviation of 3.1 gallons. Test the hypothesis at a 5% level of significance. [Go to answer]

3.
A certain population of the annual incomes of unskilled laborers has a standard deviation of $1200. A random sample of 36 such incomes results in a sample mean of $7280.
(a)
Give a point estimate for the population mean annual income.
(b)
Estimate with a 95% confidence interval. [Go to answer]

4.
A botanist wants to determine the proportion of tulip bulbs of a particular type that will bloom. He selects a random sample of 100 such bulbs and finds that 36 of them bloom.
(a)
Calculate a 90% confidence interval for this proportion.
(b)
Calculate a 98% confidence interval for this proportion. [Go to answer]

5.
By measuring the amount of time it takes a component to move from one work station to the next, an engineer has estimated that the standard deviation is 7.0 seconds. How many measurements should be taken in order to be 95% certain the maximum error of estimation will not exceed 0.8 seconds? [Go to answer]

6.
Suppose the manager of a pet supply store wants to determine if there is a difference in the amount of money spent, on the average, by owners of dogs vs. owners of cats. (Customers who own multiple pets were disregarded in this analysis.) The results for a sample of 37 dog owners were an average of $26.47, and a standard deviation of $9.45. The results for a sample of 26 cat owners were an average of $19.16, and a standard deviation of $8.52. Test the hypothesis at a 5% level of significance. [Go to answer]

7.
A corporate personnel manager is in charge of promoting the "wellness" of employees. One target is lowering blood pressure for employees who are under stress. The manager wants to test the effectiveness of a stress-reduction program designed to lower systolic blood pressure. Ten employees with high blood pressure were randomly selected. Their blood pressure was taken before and after participating in the stress-reduction program. Here is the data:
Employee  Before  After
   1        158    148
   2        176    133
   3        150    152
   4        179    170
   5        183    155
   6        206    178
   7        177    185
   8        165    151
   9        175    180
  10        186    144
At the 5% level of significance, is there evidence to suggest the program lowers blood pressure? [Go to answer]

8.
The average weekly earnings for all full-time employees in your company are reported to be $344 with a standard deviation of $110. Suppose you want to check this claim since you suspect the mean is actually higher than that. You collect a random sample of 1200 employees and find a sample mean of $361. Can you disprove the claim? Test at a 5% level of significance. [Go to answer]

9.
In an air pollution study, an experiment station obtained a mean of 2.36 micrograms suspended benzene-soluble organic matter per cubic meter with a standard deviation of 0.48 from a random sample of size 10. Construct a 99% confidence interval for the true mean of the population from which the sample was taken. [Go to answer]

10.
Suppose that the time it takes to find Waldo in a "Where's Waldo" picture has a standard deviation of 2 minutes. A random sample of 10 business students were given the same picture. The average time that it took to find Waldo was found to be 1.2 minutes. Construct a 99% confidence interval for the true mean time for a business student to find Waldo. [Go to answer]

11.
A large department store chain issues its own credit card. The research director wants to find out if the mean monthly unpaid balance is different from $400. A random sample of 45 unpaid balances reveals a mean of $407 and a standard deviation of $38. Is there evidence at $\alpha = .05$ to conclude the population mean balance is different from $400? [Go to answer]

12.
A cable TV company would like to estimate the proportion of its customers who would purchase a cable TV guide. The company would like to have 90% confidence that its estimate is correct to within $\pm 0.05$ of the true proportion. Experience in other areas indicates that 30 percent of the customers have purchased such guides in the past. What sample size is necessary here? [Go to answer]

13.
The manager of a local engine repair service is considering mailing out thousands of coupons that offer a 50% discount on engine tune-ups. However, an industry trade publication indicates that at most 20% of all such coupons will be used. The manager believes that it will only be worthwhile to mail out the coupons if the proportion is actually higher than that. To get an idea, coupons are mailed out to 250 car owners, of which 96 made use of the coupon. Is there evidence at $\alpha = .10$ to suggest a mass mailing of these coupons will be worth it to the manager? [Go to answer]

14.
To estimate the number of visitors per month to the John F. Kennedy Library, the following numbers of visitors were recorded for 12 randomly selected months:
198,286   249,821   294,653   255,728   267,475   231,759
275,641   191,374   228,391   307,613   250,834   259,540
Estimate the mean number of visitors per month with a 95% confidence interval. [Go to answer]

15.
A presidential aide said that at least half of the nation agreed with the U.S. invasion of Granada in 1984. However, a Roper poll of 500 Americans found that only 39% supported the invasion. Did the 39% happen by chance, or is there statistical evidence that the aide was in error? Use $\alpha = .05$. [Go to answer]

16.
To determine the flow characteristics of oil through a valve, the inlet oil temperature is measured in degrees Fahrenheit. Here is a sample of 12 readings:
93   99   97   99   94   91
93   90   89   92   90   93
We want to know whether there is evidence that the mean inlet oil temperature is significantly less than 98 degrees. Test at $\alpha = .01$. [Go to answer]

17.
A high-volume drug screen was designed to find compounds that reduce low-density lipoproteins (LDL) cholesterol in quail. The treatment group of quail were fed a special diet mixed with a drug compound over a specified period of time. The placebo group of quail were fed the same special diet for the same period of time but without the drug compound. Following are the plasma LDL levels for the two groups:

Treatment

40 31 50 48 152 44 74 38 81 64
Placebo
64 49 54 64 97 66 76 44 71 89
70 72 71 55 60 62 46 77 86 71
Is there evidence that the drug reduces LDL cholesterol on the average for these quail? Use $\alpha = .10$. Is there anything you notice about the data that might be affecting the results? [Go to answer]

18.
The superintendent of school district A and the superintendent of school district B want to compare the study habits of their students. Eleven students from each district were paired according to IQ and then their study habits were scored by an independent party. The results are given here:
District A     105  109  115  112  124  107  121  112  104  101  114
District B     115  103  110  125   99  121  119  106  100   97  105
Can you conclude that there is a difference on the average in the study habits of students at the two schools? Use a 1% level of significance. [Go to answer]


Answers

1.
(Section 7.2) t-interval for $\mu$
Here we have
n = 30,
$\overline{X}=2.6$, and
S = 0.9 (note that this standard deviation came from the sample!)
(a)
The point estimate is the sample mean, $\overline{X}=2.6$.
(b)
$\overline{X} \pm t_{\alpha/2}^{(n-1)} \frac{S}{\sqrt{n}}$
The t critical value here will have n - 1 = 30 - 1 = 29 degrees of freedom. We want a 99% interval:

t curve

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Use the MATH SOLVER (see the t-interval example page for more details) to get the t critical value.

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$2.6 \pm 2.7564 \frac{0.9}{\sqrt{30}}$
(2.147, 3.053)

Alternatively, we could use the TInterval function:

TI-83 screen

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2.
(Section 8.6) t-test for $\mu$
Here we have
n = 60,
$\overline{X}=11.3$, and
S = 3.1 (note that this standard deviation came from the sample.)

The hypotheses are:
$H_0: \mu \leq 11$
$H_1: \mu > 11$
$\emph{t} = \frac{\overline{X} - \mu}{S/ \sqrt{n}} =
\frac{11.3 - 11}{3.1/ \sqrt{60}} = 0.7496
\hspace{.15in} \ngtr \hspace{.15in}
t^{(59)}_{.05} = 1.6711$

t curve

TI-83 screen TI-83 screen

The test statistic does not fall in the rejection region, so do not reject H0.

The p-value is .2282, which is not less than $\alpha = .05$.

t curve   TI-83 screen

Again, we conclude: do not reject H0.

We could also use the T-Test function:

TI-83 screen

TI-83 screen TI-83 screen

3.
(Section 7.1) Z-interval for $\mu$
Here we have
n = 36,
$\overline{X}=7280$, and
$\sigma=1200$ (note that this standard deviation did NOT come from the sample.)
(a)
The point estimate is the sample mean, $\overline{X}=7280$.
(b)
$\overline{X} \pm Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

normal curve

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$7280 \pm 1.96 \frac{1200}{\sqrt{36}}$
(6888, 7672)

We can also use the ZInterval function:

TI-83 screen

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4.
(Section 7.3) Z-interval for p
Here we have
n = 100, and
$p_s = \frac{36}{100} = .36$.
(a)
$p_s \pm Z_{\alpha/2} \sqrt{\frac{p_s(1-p_s)}{n}}$

normal curve

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$.36 \pm 1.645 \sqrt{\frac{.36(.64)}{100}}$
(0.2810, 0.4390)

We could also use the 1-PropZInt function:

TI-83 screen

TI-83 screen TI-83 screen

(b)

normal curve

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$.36 \pm 2.326 \sqrt{\frac{.36(.64)}{100}}$
(0.2484, 0.4716)

Or, using 1-PropZInt:

TI-83 screen TI-83 screen

5.
(Section 7.4) Sample size determination
$n = \frac{(Z_{\alpha/2})^2 \sigma^2}{e^2} =
\frac{(1.96)^2 (7)^2}{(0.8)^2} = 294.11 \hspace{.2in}
\Longrightarrow \hspace{.2in} n = 295$

normal curve   TI-83 screen

Remember to always round up for sample sizes!

6.
(Section 9.1) Two sample t-test
Dog owners Cat owners
nD = 37 nC = 26
$\overline{X}_D = 26.47$ $\overline{X}_C = 19.16$
SD = 9.45 SC = 8.52
We are looking for a "difference", so the hypotheses will be:
$H_0: \mu_D = \mu_C$
$H_1: \mu_D \neq \mu_C$
$S^{2}_{p} = \frac{(n_D - 1)S_D^2 + (n_C - 1)S_C^2}{(n_D - 1) + (n_C - 1)}
= \frac{36(9.45)^2 + 25(8.52)^2}{36 + 25} = 82.45$

\begin{displaymath}t = \frac{(\overline{X}_D - \overline{X}_C) - 0}
{\sqrt{S_p^...
... 19.16}{\sqrt{82.45(\frac{1}{37} + \frac{1}{26})}} =
3.1459
\end{displaymath}

The degrees of freedom here are df = nD + nC - 2 = 37 + 26 - 2 = 61.

$t^{(61)}_{.025} = \pm 1.9996$

t curve   TI-83 screen

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The test statistic falls in the rejection region, so reject H0.
Let's get the p-value:

t curve   TI-83 screen

$p-value = 2(.00128) = .00256 \hspace{.15in} < \hspace{.15in}
\alpha = .05$, which means we reject H0.

We could also use the 2-SampTTest function:

TI-83 screen TI-83 screen

Select Pooled since we are using a pooled variance two sample t-test.

TI-83 screen

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7.
(Section 9.3) Paired t-test
Let's enter the before and after numbers into L1 and L2, respectively:

TI-83 screen

Now let's calculate the differences (Before - After, although we could do it the other way):

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Since we calculated the differences as Before - After, then POSITIVE differences imply that the blood pressure decreased due to the program. Thus the hypotheses should be $H_0: \mu_D \leq 0$
$H_1: \mu_D > 0$
Let's get summary statistics on the column of differences:

TI-83 screen TI-83 screen

So $\overline{D} = 15.9$, and SD = 18.628.
$t = \frac{\overline{D} - 0}{S_D / \sqrt{n}}
= \frac{15.9}{18.628 / \sqrt{10}} = 2.6992$
Our rejection region will look like this:

t curve


TI-83 screen TI-83 screen

The test statistic falls into the rejection region; t = 2.6992 > 1.8331, so we reject H0. Yes, there is evidence that the program lowers blood pressure.

p-value:

t curve   TI-83 screen

Our p-value is less than alpha, .01221 < .05, so again this leads us to the same conclusion: Reject H0.

We could use the T-Test function on our column of differences:

TI-83 screen TI-83 screen

8.
(Section 8.2) Z-test for $\mu$
Here, we have
n = 100,
$\overline{X} = 361$, and
$\sigma = 110$ (note that this standard deviation did NOT come from the sample.)

Our hypotheses will be:
$H_0: \mu \leq 344$
$H_1: \mu > 344$
Test statistic:
$Z = \frac{361 - 344}{110 / \sqrt{1200}} = 5.354$
Rejection region:

normal curve   TI-83 screen

The test statistic is definitely in the rejection region. Reject H0.

TI-83 screen

p-value = .00000004 < .05. Reject H0.
Using the Z-Test function:

TI-83 screen

TI-83 screen TI-83 screen

9.
(Section 7.2) t-interval for $\mu$
Here we have:
n = 10,
$\overline{X} = 2.36$, and
S = 0.48 (note that this standard deviation came from the sample.)

We need a 99% confidence interval:

t curve   TI-83 screen

TI-83 screen TI-83 screen

$2.36 \pm 3.2498 \frac{0.48}{\sqrt{10}}$
(1.867, 2.853)
Using the TInterval command:

TI-83 screen TI-83 screen

10.
(Section 7.1) Z-interval for $\mu$
Here we have:
n = 10,
$\overline{X} = 1.2$, and
$\sigma = 2$ (note that this standard deviation did NOT come from the sample.)

We need a 99% confidence interval:

normal curve

TI-83 screen TI-83 screen

$1.2 \pm 2.5758 \frac{2}{\sqrt{10}}$
(0, 2.829)
A negative time is impossible, so it makes more sense to write the lower bound as zero.

Using the TInterval command:

TI-83 screen TI-83 screen

11.
(Section 8.6) t-test for $\mu$
Here we have:
n = 45,
$\overline{X} = 407$, and
S = 38 (note that this standard deviation came from the sample.)

The hypotheses here will be:
$H_0: \mu = 400$
$H_1: \mu \neq 400$

Test statistic:
$t = \frac{407 - 400}{38 / \sqrt{45}} = 1.2357$
The rejection region will look like:

t curve

TI-83 screen TI-83 screen

$t = 1.2357 \ngtr t^{(44)}_{.025} = 2.0154$, so do not reject H0. No, there is no evidence of a difference.

p-value:

t curve   TI-83 screen

p-value $= 2(.11156) = .22312 \nless .05$, so do not reject H0.

With the T-Test function:

TI-83 screen TI-83 screen

12.
(Section 7.4) Sample size determination
$n = \frac{(Z_{\alpha/2})^2 p(1-p)}{e^2} =
\frac{(1.645)^2 .30(.70)}{(.05)^2} = 227.27 \hspace{.15in}
\Longrightarrow \hspace{.15in} n = 228$

normal curve   TI-83 screen

Remember to always round up for sample sizes.

13.
(Section 8.7) Z-test for p
Here, we have
n = 250, and
$p_S = \frac{96}{250} = .384$
The hypotheses will be:
$H_0: p \leq .20 \hspace{.2in} vs. \hspace{.2in}
H_1: p > .20$
Rejection region:

normal curve   TI-83 screen


\begin{displaymath}Z = \frac{p_S - p}{\sqrt{\frac{p(1-p)}{n}}}
= \frac{.384 - .20}{\sqrt{\frac{.2(.8)}{250}}} = 7.273
\end{displaymath}

The test statistic is most definitely in the rejection region. Reject H0. Yes, there is evidence that the coupon mailings will be worthwhile.

TI-83 screen

p-value $\approx 0$, so again we conclude: Reject H0.
We could also use the 1-PropZtest:

14.
(Section 7.2) t-interval for $\mu$
Let's enter the data into L1,

TI-83 screen

and then get the summary statistics.

TI-83 screen TI-83 screen

We see that
n = 12,
$\overline{X} = 250926.25$, and
S = 34773.64 (from the sample, obviously!)

Critical value:

t curve   TI-83 screen

TI-83 screen TI-83 screen

$250926.25 \pm 2.20099 \frac{34773.64}{\sqrt{12}}$
(228832, 273020)

TInterval function:

TI-83 screen TI-83 screen

15.
(Section 8.7) Z-test for p
Here, we have
n = 500, and
pS = .39.
The hypotheses will be:
$H_{0}: p \geq .50 \hspace{.2in} vs. \hspace{.2in}
H_{1}: p < .50$
Test statistic:

\begin{displaymath}Z = \frac{.39 - .50}{\sqrt{\frac{.50(.50)}{500}}} = -4.919
\end{displaymath}

Rejection region:

normal curve   TI-83 screen

The test statistic is in the rejection region. Reject H0. Yes, there is evidence that the aide made a false claim.

TI-83 screen

p-value = .0000004 < $\alpha = .05$. Reject H0.

At first, it seems like this problem isn't set up right for us to use the 1-PropZTest function:

TI-83 screen

But, since $.39 \times 500 = 195$ exactly, we can enter in 195 for X. Note that this only works if $p_S \times n$ results in a whole number.

TI-83 screen TI-83 screen

16.
(Section 8.6) t-test for $\mu$
Let's enter the data into L2 and get the descriptive statistics.

TI-83 screen TI-83 screen

Now we see that
n = 12,
$\overline{X} = 93.333$, and
S = 3.393 (from the sample, of course).

The hypotheses will be:
$H_{0}: \mu \geq 98$
$H_{1}: \mu < 98$
Test statistic:
$t = \frac{93.333 - 98}{3.393/ \sqrt{12}} = -4.764$
Rejection region:

t curve   TI-83 screen

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The test statistic is in the rejection region. Reject H0. There is evidence that the mean is less than 98 degrees.

p-value $= .000293 < \alpha = .01$. Reject H0.

T-Test function:

TI-83 screen TI-83 screen

17.
(Section 9.1) Two sample t-test
Let's enter the data into L3 and L4 and get descriptive statistics for each:

TI-83 screen

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Now we have the following information:
Treatment Placebo
nT = 10 nP = 20
$\overline{X}_T = 62.2$ $\overline{X}_P = 67.2$
ST = 35.42 SP = 13.97

If the drug reduces cholesterol, then the treatment mean should be less than the placebo mean. This is what we will try to show. So the hypotheses will be:
$H_{0}: \mu_{T} \geq \mu_{P}$
$H_{1}: \mu_{T} < \mu_{P}$
$S_p^2 = \frac{9(35.42)^2 + 19(13.97)^2}{10 + 20 - 2} = 535.69$
$t = \frac{62.2 - 67.2}{\sqrt{535.69(\frac{1}{10} + \frac{1}{20})}}
= -0.5578$
The degrees of freedom for this problem are df = 10 + 20 - 2 = 28.

t curve   TI-83 screen

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The test statistic does not fall in the rejection region. Do not reject H0. No, there is no evidence that the drug reduces LDL cholesterol.

t curve   TI-83 screen

p-value $= .29071 \nless .10$, thus we make the same conclusion- Reject H0.

There is an "outlier" in the Treatment group; the 5th value of 152 is quite large and is definitely affecting the results.

With the 2-SampTTest function, we see:

TI-83 screen

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18.
(Section 9.3) Paired t-test
Let's put the data into L5 and L6:

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Make a column next to L6 and name it "DIFF". Then calculate the District A - District B differences.

TI-83 screen

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We are looking for a "difference", which means a two-tailed test:
$H_{0}: \mu_{D} = 0$
$H_{1}: \mu_{D} \neq 0$

Let's get the descriptive statistics on the differences:

TI-83 screen TI-83 screen

We see that
$\overline{D} = 2.1818$,
SD = 11.1877, and
nD = 11.
The test statistic is
$t = \frac{2.1818}{11.1877/ \sqrt{11}} = 0.6468$
Rejection region:

t curve

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The test statistic does not fall in the rejection region. Do not reject H0. No, there is no evidence that the study habits differ.

t curve   TI-83 screen

p-value $= 2(.26616) = .53232 \nless \alpha =.01$

Using the T-Test function on the DIFF column:

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2001-03-12