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 1% level of significance.
Round to two decimal places if necessary
a. Calculate the test statistic.
z =
b. Determine the critical value(s) for the hypothesis test.
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
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