Question

The District of Columbia Opportunity Scholarship Program was created by Congress to provide tuition vouchers to low-income parents who want their child to attend a private school. In 2012–2014, a treatment group of 995 students were randomly selected to receive a tuition voucher, while a control group of 776 students did not receive vouchers. One year later, the students were given a standardized reading test. The average score of the students in the treatment group was 601.78 with a standard deviation of 52.65; the average score of the control group was 605.78 with a standard deviation of 55.73. At the α=0.01 level of significance, can we conclude that the average reading score of students who receive a tuition voucher to attend private school is lower than the average reading score of students who do not receive a tuition voucher?

- do a one tailed test

- have a null and alternative hypothesis

- test statistic and p-value

- conclusion.

Answer 1

The provided sample means are shown below:

Xˉ1​=601.78

Xˉ2​=605.78

Also, the provided sample standard deviations are:

s1​=52.65

s2​=55.73

and the sample sizes are n1​=995 and n2​=776.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ =μ2​

Ha:μ1​ < μ2​

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df=1769. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this left-tailed test is tc​=−2.328, for α=0.01 and df=1769.

The rejection region for this left-tailed test is R={t:t<−2.328}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$

t=$\frac{ 601.78 - 605.78}{\sqrt{ \frac{(995-1)52.65^2 + (776-1)55.73^2}{ 995+776-2}(\frac{1}{ 995}+\frac{1}{ 776}) } } = -1.546$

(4) Decision about the null hypothesis

Since it is observed that t=−1.546≥tc​=−2.328, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0611, and since p=0.0611≥0.01, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is less than μ2​, at the 0.01 significance level

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