Question

A lecturer is interested in estimating the proportion of students who fail quantitative methods exams. In a random sample of 120 students who sat for a quantitative methods exam in a certain year 24 failed. (a) Compute a 90% confidence interval for the proportion of students who fail quantitative methods. 2 (b) Is there sufficient evidence to indicate that the proportion of students who fail quantitative methods is higher than 0.15? Test at the 5% level of significance. (c) Can the confidence interval in (a) be used to answer (b)? Explain.

Answer 1

Solution:-

a) 90% confidence interval for the proportion of students who fail quantitative methods is C.I = ( 0.1399, 0.2601).

$C.I = \hat{p}\pm z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

C.I = 0.20 + 1.645*0.036515

C.I = 0.20 + 0.0601

C.I = ( 0.1399, 0.2601)

b) No, the proportion of students who fail quantitative methods is not higher than 0.15.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.15
Alternative hypothesis: P > 0.15

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.032596
z = (p - P) / S.D

z = 1.534

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 1.534.

Thus, the P-value = 0.063.

Interpret results. Since the P-value (0.063) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the proportion of students who fail quantitative methods is not higher than 0.15.

c) Yes, the confidence interval in (a) can be used to answer (b).

No, the proportion of students who fail quantitative methods is not higher than 0.15, because the lower limit of the 90% confidence interval is smaller than 0.15.