Question

A survey at Evashevski University showed that 75% of the student body believes that the school offers classes at appropriate times during the school day. To investigate with this feeling is shared by students specifically within the business department, the department polled a sample of 100 business students; 64 said they thought that classes were offered at appropriate times during the school day.

Use an α = .10.

A. Provide the appropriate hypothesis test criteria:

HO: p ≥ ≤ = HA: p < > ≠

B. Using the data from the sample, answer the five fill-in-the-blank questions, and make the correct hypothesis test conclusion.

Reject Ho if the test statistic of		is	> <	the critical value of
Reject Ho if the p-value of		is	<	the value of α of

Based on these results, we should:

• Reject Ho

• Accept Ho

C. What does this poll say about business students at Evashevski University, compared to their peers at the university as a whole?

Answer 1

Here we have given that,

n=number of business students =100

x: number of business students thought that classes were ordered at appropriate times during a school day =64

Now, we estimate the sample proportion as

=sample proportion =

p=populaiton proportion of the student body believes that the school offers classes at appropriate times during the school day =75%=0.75

The below mentioned necessary assumption is satisfied for this hypothesis test (one sample proportion test).

• The response is binary i.e. either success/failure and the distribution is binomial.
• The mean n*p=100*0.75=75 and the variance = n*p*(1-p)= 100**0.75*(1-0.75)=19 are greater than 10, the binomial distribution can be approximated by the normal distribution.

(A)

Claim: To check whether the proportion of the student with the school offers classes at appropriate times during the school day (75%) i.e. 0.75 within the business department.

The null and alternative hypothesis are as follows,

v/s

where p is the population proportion

This is a left one-tailed test.

Now, we can find the test statistic is as follows,

Z-statistics=

       =

         = -2.54

The test statistics is -2.54.

(B)

Now we find the P-value,

= level of significance= 0.10

p-value=P(Z < z-statistics) as this is left one tailed test

=P(Z < -2.54)

=0.0055 Using standard normal z table see the value corresponding to the z=-2.54

The p-value is 0.0055

Now we find the z-critical value,

=

= -1.28 Using EXCEL=ABS(NORMSINV(probablity =0.10))

Decision Rule:

Reject Ho if the test statistics of z is > than the critical value of z, otherwise fail to reject Ho.

and

Reject Ho if the p-value of z is < than the value of , otherwise fail to reject Ho.

Here in our case,

p-value (0.0055) < 0.10

and

z-statistics (-2.54) > z-critical (-1.28)

i.e. we reject the Null hypothesis Ho.

(C)

Conclusion:

There is sufficient evidence to support the claim the proportion of the student with the school offers classes at appropriate times during the school day (75%) i.e. 0.75 within the business department.