Question

1. Assume that we want to construct a confidence interval. Do one of the​ following, as​...

1. Assume that we want to construct a confidence interval. Do one of the​ following, as​ appropriate: (a) find the critical value t Subscript alpha divided by 2​, ​(b) find the critical value z Subscript alpha divided by 2​, or​ (c) state that neither the normal distribution nor the t distribution applies. Here are summary statistics for randomly selected weights of newborn​ girls: n=181​, x= 33.7 ​hg, s= 7.6 hg. The confidence level is 99​%.

a. tα/2e=

B. za/2 =

c. Neither the normal distribution nor the t distribution applies.

2. Here are summary statistics for randomly selected weights of newborn​ girls: n=157​, x =31.2 hg, s=6.9 hg. Construct a confidence interval estimate of the mean. Use a 90​% confidence level. Are these results very different from the confidence interval 29.6 hg <μ<32.0 hg with only 15 sample​ values, x =30.8hg, and s=2.7 hg?

-----hg<μ< ---- Round to one decimal place

3. A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 15 subjects had a mean wake time of 104.0 min. After​ treatment, the 15 subjects had a mean wake time of 76.4 min and a standard deviation of 24.8 min. Assume that the 15 sample values appear to be from a normally distributed population and construct a 99​% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 104.0 min before the​ treatment? Does the drug appear to be​ effective?

-----min<μ<-----min Round to one decimal place

1)

2) At 90% confidence level, the critical value is z0.05 = 1.645

The 90% confidence interval is

= 30.29, 32.11

= 30.3, 32.1

No, the results are not very different from the confidence interval 29.6 < < 32.0.

3) df = 15 - 1 = 14

At 99% confidence level, the critical value is t* = 2.977

The 99% confidence interval is

Yes, the drug appear to be effective, since 104 does not lie in the confidence interval.

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