The mean 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 ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than \(60\) hours?

Data (length of mean work week): \(70; 45; 55; 60; 65; 55; 55; 60; 50; 55\).

As per the information, the mean work week for engineers in a start-up company is \(60\) hours. A newly hired engineer thinks that it is shorter and asks ten engineering friends for the length of their mean work weeks. She calculated the length of the week using the following data:

\(70, 45, 55, 60, 65, 55, 60, 50, 55\)

Based on the results she counted that the mean work week is shorter than \(60\) hours. Now conduct a hypothesis testing to determine if the mean work week length is \(60\) hours or less than \(60\) hours.

Hypothesis test is usually performed on a sample data taken from larger population. This test is used to obtain results based on a hypothesis performed on the sample data. Usually the analysts perform this test to determine whether his hypothesis is true or false.

Null hypothesis states that there is no statistical connection between two variables and hence the researcher always uses this to disapprove the inference. Alternate hypothesis states that there is a significant relationship between two variables.

Student’s t-distribution is also a part of continuous probability distribution methods which is used when the standard deviation value is unknown and the sample size for any given population is more than or equal to \(30\). This method is used again when the population is normally distributed for any sample size and the values of standard deviation is unknown.

First let us determine the null and alternate hypothesis:

Null hypothesis states that the mean work week length is \(60\) hours and the alternate hypothesis states that the mean work week length is less than \(60\) hours.

\(H_{0}:\mu \geq 60\)

\(H_{a}:\mu < 60\)

Here the random variable which is \(\bar{X}\), the mean length of the work week. Since we do not know the population standard deviation we use student’s t-distribution for this test.

\(t_{n-1}=t_{10-1}\)

\(t_{9}\)

Thus, the t test statistic is calculated using the following formula:

\(t=\frac{\bar{X}-\mu}{s/\sqrt{n}}\)

Here \(\bar{X}\) is the sample mean and µ is the population mean, \(s\) is the standard deviation and n is the sample size. Now let’s calculate the sample mean using the data given in the question:

\(\bar{X}=\frac{70+45+55+60+65+55+55+60+50+55}{10}\)

\(=\frac{570}{10}\)

\(=57\)

Therefore let us calculate \(z\) value using the formula-

\(z=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\)

\(=\frac{57-60}{7.15/\sqrt{10}}\)

\(=\frac{-3}{7.15/3.162}\)

\(=\frac{-3}{2.26}\)

\(=-1.327\)

After this step let us find the \(r-\)value using NORMSDIST function in excel.

Thus the following is the student’s \(t-\)distribution curve for the hypothesis test.

We know that alpha value is \(0.05\) and the \(p-\)value is greater than the alpha value therefore the null hypothesis is not rejected. Since null hypothesis is not rejected we do not have enough evidence to state that that the the mean work week length is less than \(60\) hours.

Now let us determine the \(95%\) confidence interval:

Thus the \(95%\) confidence level indicates that the mean work week length lies within \(52.57\) and \(109.57\). Let us represent the same in the student’s t-distribution curve