A newspaper article stated that college students at a local university spent an average of $95 on beer a year. A student investigator who believed the average stated was too high polled a random sample of 50 students and found a mean of $92.25 and s = 10 Use these results to test at the .05 level of significance the statement made by the newspaper.

Use a one-sample z-test.

Hypotheses:
Ho: µ = 95 -->null hypothesis
Ha: µ < 95 -->alternate hypothesis

Formula:

z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

With your data:
z = (92.25 - 95)/(10/√50) = ?

Finish the calculation.

Check a z-table at .05 level of significance for a one-tailed test.
If the z-test statistic exceeds the critical value from the z-table, reject the null. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null. If you reject the null, the conclusion will be µ < 95.

I hope this will help get you started.

Need step by step format can't figure out, need an example:

two samples of data regarding salary of Co A and Co B determine if you can be 99% sure that Co?
A makes more than Co B? Company A - Mean 55,000, Std Dev 5,000, Sample Size 200; Co B mean 52,000, Std Dev 3,000 Sample size 400 please show steps thanks

two hundred math 115 students kept trak of the number of hours of homework they did during the fall 94 semester. the results are found to closely approximate a normal distribution with mean 14 hours and standard deviation 2 hours. determine the fraction of the students who studied more than 16 hours

To test the statement made by the newspaper, we can use a hypothesis test. The null hypothesis (H₀) will assume that the average amount spent on beer is $95, while the alternative hypothesis (H₁) will assume that the average amount spent on beer is not $95.

Let's set up the hypotheses formally:

H₀: The average amount spent on beer = $95
H₁: The average amount spent on beer ≠ $95

We are given a sample of 50 students with a sample mean of $92.25 and a sample standard deviation (s) of $10.

To perform the hypothesis test, we can use the two-sample t-test since the population standard deviation is unknown. However, since we are testing against a specified value ($95) rather than comparing two samples, it is essentially a one-sample t-test.

Now, let's calculate the test statistic, which is the t-value:

t = (sample mean - population mean) / (sample standard deviation / √n)
= (92.25 - 95) / (10 / √50)

Using a calculator or statistical software, we can calculate the t-value to be approximately -1.50.

Next, we need to compare the t-value to the critical t-value at the 0.05 level of significance.

For a two-tailed test at a 0.05 significance level (α = 0.05), we divide the significance level by 2 (0.025) and determine the critical t-value with degrees of freedom (df) equal to n - 1 (50 - 1 = 49). Looking up in the t-distribution table, the critical t-value for a two-tailed test with 49 degrees of freedom is approximately ±2.01.

Since the calculated t-value (-1.50) is not in the critical region (not falling outside the range of ±2.01), we fail to reject the null hypothesis.

Therefore, based on the given data, we do not have enough evidence to conclude that the average amount spent on beer by college students at the local university is different from $95.