Mathematics, 28.05.2020 21:58, misstaejailicious200

A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The mean and standard deviation of their scores were 75.1 and 12.8, respectively. In a random sample of 50 students who only completed high school, the mean and standard deviation of the test scores were 72.1 and 14.6, respectively We want to infer at the 10% significance level that a difference exists between the two groups. Based on the results of question 29, what is the value of test statistics

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answered: coolstacie

We conclude that no significant difference exists between the two groups.

Step-by-step explanation:

We are given that a random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test.

The mean and standard deviation of their scores were 75.1 and 12.8, respectively. In a random sample of 50 students who only completed high school, the mean and standard deviation of the test scores were 72.1 and 14.6, respectively.

Let $\mu_1$ = true mean score for undergraduate students.

$\mu_2$ = true mean score for high school students.

SO, Null Hypothesis, H_0 : $\mu_1=\mu_2$ {means that no significant difference exists between the two groups}

Alternate Hypothesis, H_A : $\mu_1\neq \mu_2$ {means that a difference exists between the two groups}

The test statistics that would be used here Two-sample t-test statistics as we don't know about population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ t__n__1-_n__2-2

where, $\bar X_1$ = sample mean score for undergraduate students = 75.1

$\bar X_2$ = sample mean score for high school students = 72.1

s_1 = sample standard deviation for undergraduate students = 12.8

s_2 = sample standard deviation for high school students = 14.6

n_1 = sample of undergraduate students = 35

n_2 = sample of high school students = 50

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 12.8^{2} +(50-1)\times 14.6^{2} }{35+50-2} }$ = 13.891

So, the test statistics = $\frac{(75.1-72.1)-(0)}{13.891 \times \sqrt{\frac{1}{35} +\frac{1}{50} } }$ ~ t_8_3

= 0.979

The value of t test statistics is 0.979.

Now, at 10% significance level the t table gives critical values of -1.666 and 1.666 at 83 degree of freedom for two-tailed test.

Since our test statistic lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that no significant difference exists between the two groups.

answered: Guest

step-by-step explanation:

because that's the answer