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 144At the 5% level of significance, is there evidence to suggest the program lowers blood pressure?
198,286 249,821 294,653 255,728 267,475 231,759 275,641 191,374 228,391 307,613 250,834 259,540Estimate the mean number of visitors per month with a 95% confidence interval.
93 99 97 99 94 91 93 90 89 92 90 93We want to know whether there is evidence that the mean inlet oil temperature is significantly less than 98 degrees. Test at .
Treatment
40 31 50 48 152 44 74 38 81 64Placebo
64 49 54 64 97 66 76 44 71 89 70 72 71 55 60 62 46 77 86 71Is there evidence that the drug reduces LDL cholesterol on the average for these quail? Use . Is there anything you notice about the data that might be affecting the results?
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 105Can 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.
Answers
Use the MATH SOLVER
(see the t-interval
example page
for more details) to get the t critical value.
(2.147, 3.053)
Alternatively, we could use the TInterval
function:
The hypotheses are:
The test statistic does not fall in the rejection region, so do not reject H_{0}.
The p-value is .2282, which is not less than
.
Again, we conclude: do not reject H_{0}.
We could also use the T-Test
function:
We can also use the ZInterval
function:
We could also use the 1-PropZInt
function:
Or, using 1-PropZInt
:
Dog owners | Cat owners |
n_{D} = 37 | n_{C} = 26 |
S_{D} = 9.45 | S_{C} = 8.52 |
The test statistic falls in the rejection region, so reject H_{0}.
Let's get the p-value:
,
which means we reject H_{0}.
We could also use the 2-SampTTest
function:
Select Pooled
since we are using a pooled variance two sample
t-test.
Now let's calculate the differences (Before - After, although we
could do it the other way):
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
Let's get summary statistics on the column of differences:
So
,
and
S_{D} = 18.628.
Our rejection region will look like this:
The test statistic falls into the rejection region;
t = 2.6992 > 1.8331, so we reject H_{0}.
Yes, there is evidence that the program lowers blood pressure.
p-value:
Our p-value is less than alpha,
.01221 < .05, so again this
leads us to the same conclusion: Reject H_{0}.
We could use the T-Test
function on our column of differences:
Our hypotheses will be:
Test statistic:
Rejection region:
The test statistic is definitely in the rejection region.
Reject H_{0}.
p-value
= .00000004 < .05. Reject H_{0}.
Using the Z-Test
function:
We need a 99% confidence interval:
(1.867, 2.853)
Using the TInterval
command:
We need a 99% confidence interval:
(0, 2.829)
A negative time is impossible, so it makes more sense to write
the lower bound as zero.
Using the TInterval
command:
The hypotheses here will be:
The rejection region will look like:
Test statistic:
,
so do not reject H_{0}.
No, there is no evidence of a difference.
p-value:
p-value , so do not reject H_{0}.
With the T-Test
function:
p-value ,
so again we conclude: Reject H_{0}.
We could also use the 1-PropZtest
:
and then get the summary statistics.
We see that
n = 12,
,
and
S = 34773.64 (from the sample, obviously!)
Critical value:
(228832, 273020)
TInterval
function:
The test statistic is in the rejection region. Reject H_{0}.
Yes, there is evidence that the aide made a false claim.
p-value = .0000004 < . Reject H_{0}.
At first, it seems like this problem isn't set up right for
us to use the 1-PropZTest
function:
But, since
exactly, we can
enter in 195 for X
. Note that this only works if
results in a whole number.
Now we see that
n = 12,
,
and
S = 3.393 (from the sample, of course).
The hypotheses will be:
Test statistic:
Rejection region:
The test statistic is in the rejection region. Reject H_{0}. There is evidence that the mean is less than 98 degrees.
p-value . Reject H_{0}.
T-Test
function:
Now we have the following information:
Treatment | Placebo |
n_{T} = 10 | n_{P} = 20 |
S_{T} = 35.42 | S_{P} = 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:
The degrees of freedom for this problem are
df = 10 + 20 - 2 = 28.
The test statistic does not fall in the rejection region.
Do not reject H_{0}. No, there is no evidence that the drug
reduces LDL cholesterol.
p-value
,
thus we make the same conclusion-
Reject H_{0}.
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:
Make a column next to L_{6} and name it "DIFF
".
Then calculate the District A - District B differences.
We are looking for a "difference", which means a two-tailed test:
Let's get the descriptive statistics on the differences:
We see that
,
S_{D} = 11.1877, and
n_{D} = 11.
The test statistic is
Rejection region:
The test statistic does not fall in the rejection region.
Do not reject H_{0}. No, there is no evidence that the study
habits differ.
p-value
Using the T-Test
function on the DIFF
column: