In an interview for a secretary position at the dealer, a typist claims a tying speed of 45 words per minute. On the basis of 70 trials, she demonstrated an average speed of 43 words per minute with a standard deviation of 15 words per minute. Test at 5% significance level on the typist’s claim.

According to the typist claim, we build an hypothesis test, find the test statistic and the p-value relative to this test statistic, reaching a conclusion that:

The p-value of the test is 0.1333 > 0.05, which means that there is not enough evidence to reject the typist's claim.

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In an interview for a secretary position at the dealer, a typist claims a tying speed of 45 words per minute.

At the null hypothesis, we test if the mean is of at least 45, that is:

At the alternative hypothesis, we test if the mean is of less than 45, that is:

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The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

In which X is the sample mean, is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

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45 is tested at the null hypothesis:

This means that

On the basis of 70 trials, she demonstrated an average speed of 43 words per minute with a standard deviation of 15 words per minute.

This means that

Value of the test statistic:

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P-value of the test and decision:

The p-value of the test is found using a left-tailed test(test if the mean is less than a value), with 70 - 1 = 69 degrees of freedom and t = -1.12.

Using a t-distribution calculator, the p-value is of 0.1333.

The p-value of the test is 0.1333 > 0.05, which means that there is not enough evidence to reject the typist's claim.

A similar problem can be found at

45

step-by-step explanation: