Question

Suppose it is desired to test the hypothesis that the mean score of students on a national examination is 500 against the alternative hypothesis that it is less than 500. A random sample of 15 students is taken from the population and produces a sample mean score of 475 and a sample standard deviation of 35. Assume the population of test scores is normally distributed. State the decision rule, the test statistic, and your decision.

Answer 1

Solution :

The most appropriate test for the given scenario is one sample t-test.

Null and alternative hypotheses :

The null and alternative hypotheses are as follows :

Where, μ_{​​​​​​} is mean score of students on a national examination.

Critical value and Decision rule :

In our question significance level is not given. Generally significance level of 0.05 or 0.01 or 0.10 is used. We shall use significance level of 0.05.

Significance level = 0.05

Degrees of freedom = (n_{​​​​​} - 1) = (15 - 1) = 14

Since, our test is left-tailed test, therefore we shall obtain left-tailed critical t value at 0.05 significance level and 14 degrees of freedom. The left-tailed critical t value is given by,

Critical value = -t_{(0.05, 14)} = -1.7613

For left-tailed test we make decision rule as follows :

If value of the test statistic is less than the left-tailed critical value, then we reject the null hypothesis (H_{​​​​​​0}). Otherwise we fail to reject the null hypothesis (H_{​​​​​​0}).

Test statistic :

d) The test statistic is given as follows :

Where, x̅ is sample mean, s is sample standard deviation, n is sample size and μ is hypothesized value of population mean under H_{​​​​​​0}.

We have,  x̅ = 475, s = 35, n = 15 and  μ = 15

The value of the test statistic is -2.7664.

Decision :

We have, Critical value = -1.7613

Test statistic = -2.7664

(-2.7664 < -1.7613)

Since value of the test statistic is less than the left-tailed critical value, therefore we shall reject the null hypothesis at 0.05 significance level.

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