Question

Results from previous studies showed 79% of all high school seniors from a certain city plan to attend college after graduation. A random sample of 200 high school seniors from this city reveals that 162 plan to attend college. Does this indicate that the percentage has increased from that of previous studies? Test at the 5% level of significance. Compute the z or t value of the sample test statistic

Answer 1

Answer with explanation:

Let p be the population proportion.

Then, Null hypothesis : $H_0 : p=0.79$

Alternative hypothesis : $H_a : p>0.79$

Since the alternative hypothesis is right tailed , then the hypothesis test is a right tailed test.

Given :  A random sample of 200 high school seniors from this city reveals that 162 plan to attend college.

i.e. n = 200 > 30 , so we use z-test (otherwise we use t-test.)

$\hat{p}=\dfrac{162}{200}=0.81$

Test statistic for population proportion :

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}$

$=\dfrac{0.81-0.79}{\sqrt{\dfrac{0.81(1-0.81)}{200}}}=0.720984093915\approx0.72$

Thus, the z-value of the sample test-statistic = 0.72

P-value for right tailed test =$P(z>0.72)=1-P(\leq0.72)$

$=1-0.7642375=0.2357625\approx0.2358$

Since the p-value is greater than the significance level (0.05) , so we fail to reject the null hypothesis.

Hence, we conclude that we do not have sufficient evidence to support the claim that the percentage has increased from that of previous studies.