Question

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.
Size of sample Average score Standard deviation
Last year 125
562 40
This year 100
568 42
Can the professor reject the null hypothesis that the average scores improved? Use a 5 % level of significance.

Answer 1

Let μ1​ : Population mean score of students this year and

μ2​ : 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 : μ1​≥μ2​ against the alternate hypothesis H₁: μ1​<μ2​

Thus, we use a two-sample Z 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,

$\bar{x_{1}} = 568, \bar{x_{2}} = 562, {n_{1}} =100, {n_{2}} = 125, \sigma_{1}^2 = 42^2, \sigma_{2}^2 = 40^2,\bar x1​​=568,\bar x2​​=562,n1​=100,n2​=125$  

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.