Question

1. From a random sample of 8^th graders, we find that the mean math score is 51.3. The standard deviation is 9.9. The sample size is 6,127. Construct a confidence interval for 95%. With 95% confidence, what is the interval that encompasses the true population mean? Answer this in one sentence.

Using the same example as above, can we say with 95% confidence (using a two- tailed test) that the mean math score in the population is above 50? Test this hypothesis. Compare the hypothesis test with the results from your confidence interval. Are they consistent?

From this same survey, we find the mean math score is 50.8 for girls and 51.8 for boys. The standard deviation for girls is 9.5 and the standard deviation for boys is 10.3. There are 3,129 girls in the sample and 2,998 boys. Test the hypothesis that the mean for boys and the mean for girls are different at the .05 alpha level (95% confidence) using a two-tailed test. Use df=6,028. 

Answer 1

$1.\bar{x}=51.3,\sigma=9.9,n=6127,\alpha=0.05$

$Z_{0.05}=2.645$ ,

95% confidence interval is given by-

$=\mu\pm Z_{0.05}\times \dfrac{9.9}{\sqrt{6127}}\\[9pt]=51.3\pm 2.645\times 0.1264\\[9pt]=(51.3\pm 0.334)=(50.96,51.63)$

Let $H_o: \mu>50$ , The mean math score in the population is above 50.

Test-statics is-

$z=\dfrac{u-\bar{x}}{\sigma} \\[9pt] =\dfrac{50-51.3}{9.9} \\[9pt] =\dfrac{-1.3}{9.9}=-0.131$

P-value-

$P(z>-0.131)=0.5524$

Here p-value is greater than $\alpha$ i.e. 0.05, so $H_o$ is accepted. The mean math score in the population is above 50.

Let H_o: $\mu_1=\mu_2\text{ or }\mu_1-\mu_2= 0$ , Enough evidence to support the claim that the mean for boys and the mean for girls are different at 95% confidence level.

$H_a:\mu_1\neq \mu_2$ , not having Enough evidence to support the claim that the mean for boys and the mean for girls are different at 95% confidence level.

Here, $\bar{x_1}=50.8,\bar{x_2}=51.8,s_1=9.5,s_2=10.3 ,n_1=3129,n_2=2998$

Test statics is-

$z=\dfrac{\bar{x_1}-\bar{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\[9pt]=\dfrac{50.8-51.8}{\sqrt{\frac{(9.5(^2}{3129}+\frac{(10.3)^2}{2998}}}\\[9pt]=\dfrac{1}{\sqrt{0.02884+0.0353}}\\[9pt]=\dfrac{1}{\sqrt{0.06422}}=3.952$

p-value is-

$P(z>3.952)=0.476$

Conclusion: As P-value is greaer than $\alpha$ i.e. 0.05, so Null hypothesis is accepted i.e. There is Enough evidence to support the claim that the mean for boys and the mean for girls are different at 95% confidence level.