Question

a). Professor Jennings claims that only 35% of the students at Flora College work while attending school. Dean Renata thinks that the professor has underestimated the number of students with part-time or full-time jobs. A random sample of 79 students shows that 35 have jobs. Do the data indicate that more than 35% of the students have jobs? Use a 5% level of significance.

What is the value of the sample test statistic? (Round your answer to two decimal places.)____?

b). A machine in the student lounge dispenses coffee. The average cup of coffee is supposed to contain 7.0 ounces. A random sample of six cups of coffee from this machine show the average content to be 7.3 ounces with a standard deviation of 0.70 ounce. Do you think that the machine has slipped out of adjustment and that the average amount of coffee per cup is different from 7 ounces? Use a 5% level of significance.

What is the value of the sample test statistic? (Round your answer to three decimal places.) ___________?

Answer 1

a) Let the true proportion of students having job be P.

Hypothesis can be framed as -

Null hypothesis : P = 0.35

Alternative hypothesis : P > 0.35

Test statistic is given by -

where, n = sample size = 79

p = sample proportion = 35/79 = 0.44

Po = specified value of P under the null hypothesis = 0.35

Hence, the value of the test statistic will be -

= 1.66

The critical value of z for one tailed test at 0.05 level of significance is 1.645 (as obtained from the z table by finding the z corresponding to the area approximately equal to 0.05).

As the value of test statistic > critical value of z, the null hypothesis may be rejected at 5% level of significance, hence, true proportion of students having job is greater than 0.35.

b) Null hypothesis : The average cup of coffee contains 7 ounces of coffee.

Alternative hypothesis : Average content of coffee is different from 7 ounces.

Test statistic is given by -

where, is the average coffee content in the sample = 7.3 ounces

is the hypothesed value of average coffee content under the null hypothesis = 7 ounces

s is the sample standard deviation = 0.70 ounces

n is the sample size = 6

Hence, value of the test statistic will be -

= 1.049

Degrees of freedom = n - 1 = 6 - 1 = 5

The critical value of t with 5 degrees of freedom at 0.05 level of significance = 2.571(can be obtained from the t table corresponding to 5 degrees of freedom and 0.025 probability, since it is two tailed test).

Since, the value of test statistic < critical value of t, we may Reject the null hypothesis at 5% level of significance, hence, the average content of the coffee is different from 7 ounces.