Find the p value

Question

The average work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it's shorter. She asks 10 engineering friends in start-ups for the lengths of their average work weeks. Based on the results that follow, should she count on the average work week to be shorter than 60 hours? Conduct a hypothesis test at the 5% level.

Data (length of average work week): 70; 50; 60; 65; 65; 55; 55; 55; 55; 55.

Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

What is the p-value? (Round your answer to four decimal places.)

Answer 1

Hypothesis Test:

Null Hypothesis: The average work week is 60 hours

Alternative Hypothesis: The average work week is less than 60 hours

Test Statistic: t-test

Data Set: 70; 50; 60; 65; 65; 55; 55; 55; 55; 55

Degrees of Freedom (DF): 10-1 = 9

Using the calculated t-test statistic and the t-table with α = 0.05 and DF = 9, the critical value = -2.262.

Calculate t-test statistic:

Compute the sample mean (x¯):

x¯ = (70 + 50 + 60 + 65 + 65 + 55 + 55 +55 + 55 + 55)/10 = 59

Compute the sample standard deviation (s):

s = sqrt(((70-59)^2 + (50-59)^2 + (60-59)^2 + (65-59)^2 + (65-59)^2 + (55-59)^2 + (55-59)^2 + (55-59)^2 + (55-59)^2 + (55-59)^2)/(9)) = 7.214

Compute t-test statistic:

t-test statistic = (59-60)/(7.214/sqrt(10)) = -0.140

Conclusion:

Since the calculated t-test statistic (-0.14) does not exceed the critical value (-2.262) and is not in the critical region, we fail to reject the null hypothesis. Therefore, she should count on the average work week to be about 60 hours at the 5% level.

p-value = 0.9087

Answer 2

To start, we need to define our Null and Alternative Hypotheses. I’ve written them in words below:

Null Hypothesis (H₀): The average work week for engineers in start-up companies is 60 hours or longer.

Alternative Hypothesis (H_a): The average work week for engineers in start-up companies is shorter than 60 hours.

We will use a one-sample t-test here to test these hypotheses. Let's calculate the p-value based on the provided data:

Length of Average Work Week: 70; 50; 60; 65; 65; 55; 55; 55; 55; 55.

Step 1: We first need to calculate the t-test statistic. The formula for the test statistic is given as follows:

t = (x̄ - μ) / (s / √n)

Based on this formula, we only have two given variables, which are μ (60) and n (10). We need to find two additional variables, (x̄) and (s), which represent the sample mean and sample standard deviation, respectively.

The sample mean can be calculated by taking the sum of all the data values and dividing it by the total number of observations, as follows:

x̄ = (70 + 50 + 60 + 65 + 65 + 55 + 55 + 55 + 55 + 55) / 10

x̄ = 580 / 10

x̄ = 58 hours

To find the sample standard deviation, we use the formula:

s = [√∑(x_i - x̄)² / (n - 1)]

And substitute values in for x_i, x̄, and n:

s = √[(70 - 58)² + (50 - 58)² + (60 - 58)² + (65 - 58)² + (65 - 58)² + (55 - 58)² + (55 - 58)² + (55 - 58)² + (55 - 58)² + (55 - 58)² / (10 - 1)]

Simplifying:

s = √[62 + 64 + 4 + 49 + 49 + 9 + 9 + 9 + 9 + 9 / 9]

s = √30.67

s = 5.53 hours

Plugging in our values for the sample mean and sample standard deviation into the t-test statistic equation, we get:

t = (58 - 60) / (5.53 / √10)

Solving for t, we get:

t = -2.04

Step 2: Calculate the p-value

Using the t-table or statistical software both work in finding the p-value. In this problem, the p-value for the t-statistic of -2.04, with 9 degrees of freedom (10 - 1), is approximately 0.0358.

Here is a formal conclusion to finish out the problem:

Because the p-value of 0.0358 is less than the significance level of 0.05, we reject the null hypothesis. The data provides sufficient evidence suggesting that the average work week for engineers in start-up companies is shorter than 60 hours.