The average typing speed of a good typist is said to be 60 words per minute. In a certain company, 40 applicants were given a typing test and the results showed a mean of 55 words per minute with a standard deviation of 10 minutes. Based on the results, can we say that the typing speed of the 45 applicants is below the average at the 0.05 level? Solution: 1. Formulate the null and alternative hypothesis. H:\ H: 2. Set the test statistic and level of significance. 3. Decision Rule Reject H0 if 4. Computation Given:
The Correct Answer and Explanation is:
To determine whether the typing speed of the applicants is significantly below the average of 60 words per minute, we perform a hypothesis test using the following steps:
1. Formulate the null and alternative hypothesis
- Null hypothesis (H₀): μ = 60
(The mean typing speed of applicants is equal to the population average.) - Alternative hypothesis (H₁): μ < 60
(The mean typing speed of applicants is less than the population average.)
2. Set the test statistic and level of significance
- This is a one-tailed t-test for the mean because the population standard deviation is not known, and the sample size is relatively small (n = 40).
- Significance level: α = 0.05
3. Decision Rule
- Use the t-distribution with degrees of freedom:
df = n – 1 = 40 – 1 = 39 - Find the critical t-value for a one-tailed test at α = 0.05 and df = 39.
From the t-distribution table:
t₀.05,39 ≈ -1.685 - Reject H₀ if the computed t-value is less than -1.685
4. Computation
Given:
- Sample mean (x̄) = 55
- Population mean (μ) = 60
- Standard deviation (s) = 10
- Sample size (n) = 40
Compute the test statistic (t):t=xˉ−μs/n=55−6010/40=−510/6.3246=−51.5811≈−3.16t = \frac{\bar{x} – \mu}{s / \sqrt{n}} = \frac{55 – 60}{10 / \sqrt{40}} = \frac{-5}{10 / 6.3246} = \frac{-5}{1.5811} ≈ -3.16t=s/nxˉ−μ=10/4055−60=10/6.3246−5=1.5811−5≈−3.16
5. Decision
Since -3.16 < -1.685, we reject the null hypothesis.
6. Conclusion
At the 0.05 level of significance, there is sufficient evidence to conclude that the typing speed of the applicants is significantly below the average of 60 words per minute.
Explanation
In statistical testing, we often compare a sample to a known or assumed population mean to assess if there is a significant difference. Here, we are evaluating if a group of 40 applicants has a lower average typing speed than the commonly accepted value of 60 words per minute.
The first step is to set the hypotheses. The null hypothesis assumes there is no difference, while the alternative suggests the applicants perform worse. A one-tailed t-test is appropriate because the claim is directional — we are only interested in whether the mean is lower.
The standard deviation is known only from the sample, not from the population, and the sample size is moderately large, so we use a t-distribution. After identifying the degrees of freedom (n – 1 = 39), we look up the critical value that corresponds to the 0.05 level. This helps establish the threshold for rejecting the null hypothesis.
By plugging values into the t-statistic formula, we find the observed t is about -3.16. This is much more negative than the critical value of -1.685, showing that the sample mean is significantly less than 60. Because the computed t lies in the rejection region, we reject the null hypothesis.
The conclusion is that the applicants’ average typing speed is significantly lower than 60 words per minute. This finding could be useful for the company in evaluating their hiring or training processes.
