1. Homework due Wed2. Project Proposal due FridaySubmit half-page document: • Describe the question that you want to investigate. i.e population parameter of interest.• Describe the data that you obtained. If found online, provide the link

3. Quiz on Friday

Last time: Hypothesis testing for mean when population variance is known

Today: Hypothesis testing for mean when population variance is unknown

Review example:

State the null and alternative hypothesisa) H0: = 4, H1: ≠ 4b) H0: = 4, H1: > 4c) H0: p = 4, H1: p ≠ 4d) H0: p = 4, H1: p > 4

Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient’s body, but the battery pack needs to be recharged about every 4 hours. We want to know if the mean battery life exceeds 4 hours. Exceeds 4 hours

Review example:

Which picture corresponds to our test?

H0: = 4, H1: > 4

A) B) C)

Review example:

A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation = 0.2 hour.The standard error for the sample mean can be calculated by

a) 0.2 b) 0.2 /

H0: = 4, H1: > 4

SEn

Review example:

A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation = 0.2 hour.What is the distribution of test statistic under the null

a)

b) 0.2 /

H0: = 4, H1: > 4

sample statistic null value

SEz

From H0From data

Review example:

A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 4.05 hours. Assume that battery life is normally distributed with standard deviation = 0.2 hour.Compute the test statistic

a) = -1.77b) = 1.77

H0: = 4, H1: > 4

sample statistic null value

SEz

Review example:

Find p-value for the computed test statistic = 1.77

a) 0.038b) 0.962

H0: = 4, H1: > 4

Review example:

For p-value = 0.038. Do we reject or fail to reject the null, under significance level of = 0.05.

a) Rejectb) Fail to reject

H0: = 4, H1: > 4

p-value < Reject H0

p-value > Do not Reject H0

Review example:

We can construct the rejection region to reach the same conclusion Rejection region is a) (1.65, infinity )b) (0.9505, infinity )

H0: = 4, H1: > 4Alpha = 0.05

Since = 1.77 falls in the rejection region. We reject the null.

Review example:

We can also construct the confidence interval. Note here is one-sided.Lower bound = LB = 4.05 – 1.65*CI = (4.003, )

H0: = 4, H1: > 4

• With 95% confidence level, the population mean lies in the confident interval• Null value is not in the 95% confidence interval. We can reject the null hypothesis