Posted: May 1st, 2022

BUSS3104

1.

A health middle is excited about discovering a 98% confidence interval for the imply variety of days per week that Individuals who’re members of a health membership go to their health middle. Data of 215 members had been checked out and their imply variety of visits per week was 2.5 and the usual deviation was 1.9. Spherical solutions to three decimal locations the place attainable.
a. To compute the arrogance interval use a ? t z distribution.

b. With 98% confidence the inhabitants imply variety of visits per week is between and visits.

c. If many teams of 215 randomly chosen members are studied, then a special confidence interval can be produced from every group. About p.c of those confidence intervals will include the true inhabitants imply variety of visits per week and about p.c won’t include the true inhabitants imply variety of visits per week.

2.

You have an interest find a 98% confidence interval for the typical commute that non-residential college students must their faculty. The information beneath present the variety of commute miles for 13 randomly chosen non-residential faculty college students. Spherical solutions to three decimal locations the place attainable.

1524272411132713202324623

a. To compute the arrogance interval use a ? t z distribution.

b. With 98% confidence the inhabitants imply commute for non-residential faculty college students is between and miles.

c. If many teams of 13 randomly chosen non-residential faculty college students are surveyed, then a special confidence interval can be produced from every group. About p.c of those confidence intervals will include the true inhabitants imply variety of commute miles and about p.c won’t include the true inhabitants imply variety of commute miles.

three. Isabella needs to promote what number of chocolate chips are in every Huge Chip cookie at her bakery. She randomly selects a pattern of 42 cookies and finds that the variety of chocolate chips per cookie within the pattern has a imply of 11.6 and a normal deviation of 1.four. What’s the 95% confidence interval for the variety of chocolate chips per cookie for Huge Chip cookies? Assume the information is from a usually distributed inhabitants. Spherical solutions to three decimal locations the place attainable.
< μμ <

four.

You should estimate the imply temperature (in levels Fahrenheit) with the next pattern temperatures:
38.324.320.745.416.926.627.727.331.319.429.three
Discover the 80% confidence interval. Enter your reply as an open-interval (i.e., parentheses) correct to three decimal locations. Assume the information is from a usually distributed inhabitants.
80% C.I. =

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1.

A health middle is searching for a 98 p.c confidence interval for the typical variety of days per week that Individuals who’re members of a health membership go to the power. A complete of 215 members’ data had been examined, with a imply variety of visits per week of two.5 and a normal deviation of 1.9. If attainable, spherical your solutions to a few decimal locations.

a. Use a? t z distribution to compute the arrogance interval.

b. The inhabitants imply variety of visits per week is between and visits, with 98 p.c confidence.

c. If many teams of 215 randomly chosen members are studied, every group will produce a special confidence interval. The true inhabitants will probably be present in a couple of quarter of those confidence intervals.

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