A telephone company's records indicate that private customers pay on average $17.10 per month for long-distance telephone calls. A random sample of 10 customers' bills during a given month produced a sample mean of $22.10 expended for long-distance calls and a sample variance of 45. A 5% significance test is to be performed to determine if the mean level of billing for long distance calls per month is in excess of $17.10. The calculated value of the test statistic and the critical value respectively are:
(2.36, 1.8331)
(1.17, 2.2622)
(2.36, 2.2622)
(1.17, 1.8331)
(0.025, 1.8125)
Remember that standard deviation is the square root of the variance.
Formula:
t = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
t = (22.10 - 17.10)/(6.7082/√10)
I'll let you finish the calculation.
The test is one-tailed at .05 level of significance. Look at the appropriate table and determine the critical value.
Hint: degrees of freedom = n - 1 = 9
I'm sorry, I don't have the answer to that. But speaking of telephone bills, why did the scarecrow become a successful telemarketer? Because he was outstanding in his field!
To determine the test statistic and critical value, we need to perform a one-sample t-test.
Let's calculate the test statistic first.
Given:
Sample mean (x̄) = $22.10
Population mean (μ) = $17.10
Sample size (n) = 10
Sample variance (s²) = 45
The test statistic (t-score) can be calculated using the formula:
t = (x̄ - μ) / (s / √n)
Substituting the given values:
t = (22.10 - 17.10) / (√45 / √10)
= 5 / (6.708 / 3.162)
≈ 2.36
Therefore, the calculated value of the test statistic is approximately 2.36.
Now, let's determine the critical value. Since the problem states a 5% significance level, we need to find the critical value at α = 0.05 for a one-tailed test.
Looking up the critical value in the t-distribution table or using a calculator, the critical value with 9 degrees of freedom is approximately 1.8331.
Therefore, the calculated value of the test statistic and the critical value respectively are:
(2.36, 1.8331)
To determine if the mean level of billing for long-distance calls per month is in excess of $17.10, we can perform a one-sample t-test.
Let's calculate the test statistic first.
The test statistic for a one-sample t-test can be calculated as:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Given:
Sample mean (x̄) = $22.10
Population mean (μ) = $17.10
Sample size (n) = 10
Sample variance (s^2) = 45
We need to calculate the sample standard deviation (s) by taking the square root of the sample variance:
s = sqrt(sample variance) = sqrt(45) = 6.71
Now we can substitute the values into the formula:
t = (22.10 - 17.10) / (6.71 / sqrt(10))
Simplify the equation:
t = 5 / (6.71 / sqrt(10))
t = 5 * sqrt(10) / 6.71
t ≈ 2.356
The calculated value of the test statistic is approximately 2.356.
Now let's find the critical value for a 5% significance level with 9 degrees of freedom (n-1):
Using a t-table or calculator, the critical value for a one-tailed t-test with a significance level of 5% and 9 degrees of freedom is approximately 1.8331.
Therefore, the calculated value of the test statistic is approximately 2.356, and the critical value is approximately 1.8331.
The correct answer is:
(2.36, 1.8331)