Answer to Question #130044 in Statistics and Probability for Ntokozo

Question

2.1 Assume that the college professor obtained a grant to study whether there was an improvement
in the mathematics ability of freshmen. Using this grant money, he is able to obtain larger samples.
The results are compiled in the following table.
Last year
This year
Size of sample
125
100
Average score
562
568
Standard deviation
40
42
Can the professor reject the null hypothesis that the average scores improved? Use a 5 % level of
significance.

2.2 In a recent poll, 800 voters were randomly sampled and asked whether they would approve of
the building of an incinerator in the state. According to the poll, 450 voters approved and 350
voters disapproved. Is there enough evidence to argue that more than 50 % of the voters support
the building of the incinerator?
Use a 5 % level of significance.

Answer 1

Solution for 2.1:

Let $\mu_{1}$ : Population mean score of students this year and $\mu_{2}$ : Population mean score of students last year.

Similarly, let $\bar{x_{1}}$ : Sample mean score of students this year and $\bar{x_{2}}$ : Sample mean score of students last year.

Since we are asked whether the professor can reject the null hypothesis that the average score improved - H₀ : $\mu_{1} \geq \mu_{2}$ against H₁: $\mu_{1}<\mu_{2}$

Thus, we use a two-sample test for population means.

Under H₀, the test statistic is -

$Z_{calc} = \frac{\bar{x_{1}} - \bar{x_{2}}}{\sqrt{(\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}})}} \sim N(0,1)$

n₁ and n₂ are the corresponding sample sizes.

We have $\bar{x_{1}} = 568, \bar{x_{2}} = 562, {n_{1}} =100, {n_{2}} = 125, \sigma_{1}^2 = 42^2, \sigma_{2}^2 = 40^2$

Now,

$Z_{calc} = \frac{568 - 562}{\sqrt{(\frac{42^2}{100}+\frac{40^2}{125})}} = 1.087499$

We have to test the claim at level of significance of 5% (Thus alpha = 0.05). Since its a left-tailed test, the critical value from standard normal tables is:

$Z_{1-\alpha} = Z_{0.95} = -1.6449$

Test Criteria: Reject H₀ if Z_calc < Z_1-alpha (i.e. -1.6449)

Conclusion: Since our Z_calc = 1.09 is greater than -1.6449 we do not have sufficient evidence to reject H₀ at 5% level of significance.

Hence, the professor cannot reject the null hypothesis that the average scores improved.

Solution for 2.2:

Let P: Population proportion of voters who support the building of the incinerator

Similarly, let $\hat{p}$ : Sample proportion of of voters who support the building of the incinerator

Since we have to test whether there is enough evidence to argue that more than 50% of the voters support the building of the incinerator, we can set up the null and alternative hypothesis as, H₀ : $P \geq 0.50$ against H₁: $P < 0.50$

Thus, we use a one-sample test for population proportion.

Under H₀, the test statistic is -

$Z_{calc} = \frac{\hat{p} - P}{\sqrt{(\frac{P(1-P)}{n})}} \sim N(0,1)$

where n is the sample size.

And $\hat{p} = \frac{No.\ of\ voters\ in\ support}{Total\ no.\ of\ voters} = \frac{450}{800} = 0.5625$

We have n = 800

Now,

$Z_{calc} = \frac{0.5625 - 0.50}{\sqrt{(\frac{0.50*(1-0.50)}{800})}}=3.5355$

We have to test the claim at level of significance of 5% (Thus alpha = 0.05). Since its a left-tailed test, the critical value from standard normal tables is:

$Z_{1-\alpha} = Z_{0.95} = -1.6449$

Test Criteria: Reject H₀ if Z_calc < Z_1-alpha (i.e. -1.6449)

Conclusion: Since our Z_calc = 3.5355 is greater than -1.6449 we do not have sufficient evidence to reject H₀ at 5% level of significance.

Hence, there is enough evidence to argue that more than 50% voters support the building of the incinerator.