exercise: The average number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey 8 employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let X = the number of sick days they took for the past year. Should the personnel team believe that the average number is about 10?
Find the mean first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to your Z score.
4556
To determine whether the personnel team should believe that the average number of sick days is about 10, we can perform a hypothesis test. In this case, the null hypothesis (H0) is that the average number of sick days is 10, while the alternative hypothesis (Ha) is that the average is different from 10.
The steps for the hypothesis test are as follows:
Step 1: State the hypotheses.
H0: The average number of sick days is 10.
Ha: The average number of sick days is different from 10.
Step 2: Calculate the sample mean (x̄) and sample standard deviation (s) of the data.
x̄ = (12 + 4 + 15 + 3 + 11 + 8 + 6 + 8) / 8 = 8.875
s = √[( (12 - 8.875)^2 + (4 - 8.875)^2 + (15 - 8.875)^2 + (3 - 8.875)^2 + (11 - 8.875)^2 + (8 - 8.875)^2 + (6 - 8.875)^2 + (8 - 8.875)^2) / (8 - 1)]
Step 3: Set the significance level (α). This value determines the threshold for accepting or rejecting the null hypothesis. Let's assume α = 0.05 (5%).
Step 4: Compute the test statistic.
In this case, we will be using a t-test because the sample size is small and the standard deviation of the population is unknown. The formula for the t-test statistic is:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the population mean under the null hypothesis, s is the sample standard deviation, and n is the sample size.
Substituting the values:
t = (8.875 - 10) / (s / √8)
Step 5: Determine the critical value(s).
Since the alternative hypothesis is two-tailed (the average can be higher or lower than 10), we need to find the critical values in both tails of the t-distribution. To do this, we need to determine the degrees of freedom (df), which is equal to the sample size minus 1 (df = n - 1). In this case, df = 8 - 1 = 7.
Using a t-distribution table or a statistical software, we find that for a two-tailed test with α = 0.05 and df = 7, the critical values are approximately ± 2.3646.
Step 6: Compare the test statistic with the critical value(s) and make a decision.
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Step 7: State the conclusion in the context of the problem.
If we reject the null hypothesis, we conclude that there is enough evidence to suggest that the average number of sick days is different from 10. If we fail to reject the null hypothesis, we do not have enough evidence to suggest a difference.
Now, let's calculate the test statistic:
t = (8.875 - 10) / (s / √8)
We have already calculated the sample mean (x̄) and the sample standard deviation (s), so we can substitute these values.
t = (8.875 - 10) / (s / √8)
Since we do not have the actual values of s, we cannot calculate the test statistic. Could you please provide the values of the observations in the dataset?
To determine whether the personnel team should believe that the average number of sick days an employee takes is about 10, we can use hypothesis testing.
Hypothesis testing involves setting up two hypotheses: the null hypothesis (H₀) and the alternative hypothesis (H₁).
In this case, the null hypothesis (H₀) is that the average number of sick days an employee takes is 10. The alternative hypothesis (H₁) is that the average number of sick days is different from 10.
We can use a t-test to determine whether the sample data supports the null or alternative hypothesis.
To calculate the t-value and compare it to the critical value, we need to calculate the mean and standard deviation of the sample.
Mean (x̄) = (12 + 4 + 15 + 3 + 11 + 8 + 6 + 8) / 8 = 9.25
Standard Deviation (s) = square root of [(12-9.25)² + (4-9.25)² + (15-9.25)² + (3-9.25)² + (11-9.25)² + (8-9.25)² + (6-9.25)² + (8-9.25)²] / (8-1) = 3.36
Next, we calculate the t-value using the formula:
t = (x̄ - μ) / (s / sqrt(n))
Where:
x̄ = sample mean
μ = population mean (hypothesized value, i.e., 10)
s = sample standard deviation
n = sample size
Plugging in the values:
t = (9.25 - 10) / (3.36 / sqrt(8))
t = -0.75 / (3.36 / 2.83) = -0.75 / 1.186 = -0.632
Now, we compare the calculated t-value with the critical t-value. The critical t-value depends on the significance level (α) and the degrees of freedom (df). For a two-tailed test with a significance level of 0.05 and df = n-1 = 8-1 = 7, the critical t-value is approximately ±2.365.
Since the calculated t-value (-0.632) does not exceed the critical t-value (±2.365), we fail to reject the null hypothesis and do not have sufficient evidence to believe that the average number of sick days is different from 10.
Therefore, based on the given sample data, the personnel team should believe that the average number of sick days an employee takes per year is about 10.