Question

A manager at a local discount gym believes that less than 20% of gym members use the gym, at least 5 days a week. She randomly selects 100 gym members and tracks (using the electronic login system at the door) how many days they used the gym over the 2-week period. The following are the results:

2	3	10	4	2	3	8	4	8	10
5	0	6	3	9	13	6	3	12	5
3	3	5	1	5	9	8	5	8	2
6	4	4	2	12	1	3	3	2	12
7	3	14	2	8	5	2	6	1	5
6	9	6	8	10	1	11	3	2	1
5	4	1	2	3	13	7	4	8	3
7	4	3	2	10	3	1	7	11	8
4	7	6	7	8	11	7	6	3	2
5	0	4	6	5	12	2	10	1	2

Test the manager's claim at the 10% level of significance.

Standard Normal Distribution Table

a. Calculate the test statistic.

z=z=

Round to two decimal places if necessary

Enter 0 if normal approximation to the binomial cannot be used

b. Determine the critical value(s) for the hypothesis test.

• +

Round to two decimal places if necessary

Enter 0 if normal approximation to the binomial cannot be used

c. Conclude whether to reject the null hypothesis or not based on the test statistic.

Reject

Fail to Reject

Cannot Use Normal Approximation to Binomial

Please provide correct answers thanks

Answer 1

Solution:-

Given that

A manager at a local discount gym believes that less than 20% of gym members use the gym, at least 5 days a week. She randomly selects 100 gym members and tracks (using the electronic login system at the door) how many days they used the gym over the 2-week period.

Using gym for at least 5 days or week

means using gym over the 2 week period for atleast 10 days.

i.e., = sample proportion

Also test

vs

p : as proportion of gym members who use gym for atleast 5 days a week rather for atleast 10 days in 2 week period

a) Calculate the test statistic.

n = sample size

= -1.25

Z = -1.25

Reject if

i.e.,

b)  Determine the critical value(s) for the hypothesis test.

critical value : = -1.28

c)  Conclude whether to reject the null hypothesis or not based on the test statistic.

Decision :-

Since

we fail to reject

Fail to reject

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