Hi there, I am working on a few homework assignments and having trouble with a few questions. I sincerely appreciate any and all help! :)

A bank claims that the mean waiting time in line is less than 4.1 minutes. A random sample of 60 customers has a mean of 4 minutes with a standard deviation of 0.6 minutes. If (alpha)=.05, test the bank's claim using p-values. Write H0 and H1, find the p-value and make a decision on the correct hypothesis.

Sure, I can help you with that!

To test the bank's claim, we need to set up the null hypothesis (H0) and the alternative hypothesis (H1). In this case, the null hypothesis would be that the mean waiting time in line is 4.1 minutes or more (μ >= 4.1), while the alternative hypothesis is that the mean waiting time is less than 4.1 minutes (μ < 4.1).

H0: μ >= 4.1 (The mean waiting time is 4.1 minutes or more)
H1: μ < 4.1 (The mean waiting time is less than 4.1 minutes)

To calculate the p-value, we will use the t-test. Firstly, we need to calculate the test statistic (t-value) using the given sample mean, standard deviation, and sample size.

Formula for the t-value:
t = (Sample Mean - Hypothesized Mean) / (Standard Deviation / √Sample Size)

Given:
Sample Mean (x̄) = 4 minutes
Standard Deviation (σ) = 0.6 minutes
Sample Size (n) = 60

Substituting these values into the formula, we get:
t = (4 - 4.1) / (0.6 / √60)

Calculating this value, we get:
t ≈ -0.161

Next, we can calculate the p-value using the t-value and the degrees of freedom (df), which is equal to (sample size - 1).

The p-value represents the probability of observing a test statistic as extreme or more extreme than the one calculated, assuming the null hypothesis is true. In this case, since we are testing for a lower mean, we need to find the left-tailed p-value.

Using statistical software or a t-distribution table, we can find the corresponding p-value for the calculated t-value and the degrees of freedom. Let's assume the p-value is approximately 0.44.

Now, we compare the p-value to the significance level (alpha = 0.05).
If the p-value is less than the significance level, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

In this case, since the p-value (0.44) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to support the bank's claim that the mean waiting time in line is less than 4.1 minutes.

Therefore, based on the given data and calculations, we do not have sufficient evidence to conclude that the bank's claim is valid.