a. The standard deviation s = b. t test statistic = c. At the 0.05 level, the critical value te = d. Conclusion

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Chapter1: Starting With Matlab
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A drug manufacturer claims that a new drug lowers cholesterol level in patient studied. The following data
shows the levels of cholesterol in 5 patients before and after taking the drug. Assume the data is normally
distributed. Is there enough evidence to support the claim at a 0.05 level of significance?
Complete the table, then find the standard deviation of the difference s
1
3
4.
First Attempt
200
235
211
219
209
Second Attempt
195
235
194
218
204
Difference d
Mean d
(d – d)
Round all values to 3 decimal places
a. The standard deviation s=
b. t test statistic =
C. At the 0.05 level, the critical value t. =
d. Conclusion
O The sample data support the original claim
C There is not sufficient evidence to warrant rejection of the original claim
O There is sufficient evidence to warrant the rejection of the original claim
O There is not sufficient sample evidence to support the original claim
Transcribed Image Text:A drug manufacturer claims that a new drug lowers cholesterol level in patient studied. The following data shows the levels of cholesterol in 5 patients before and after taking the drug. Assume the data is normally distributed. Is there enough evidence to support the claim at a 0.05 level of significance? Complete the table, then find the standard deviation of the difference s 1 3 4. First Attempt 200 235 211 219 209 Second Attempt 195 235 194 218 204 Difference d Mean d (d – d) Round all values to 3 decimal places a. The standard deviation s= b. t test statistic = C. At the 0.05 level, the critical value t. = d. Conclusion O The sample data support the original claim C There is not sufficient evidence to warrant rejection of the original claim O There is sufficient evidence to warrant the rejection of the original claim O There is not sufficient sample evidence to support the original claim
Expert Solution
Step 1

Given that

Sample size n = 5

Level of significance = 0.05

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