A large furniture store has begun a new ad campaign on local television. Before the campaign, the long term mean daily sales were $24,819. A random sample of 40 days during the new ad campaign gave a sample mean daily sale of $25,910. Does this indicate that the population mean daily sales is now more than $24,819? Use a 1% level of significance. Assume o = $1,917.

Question

A large furniture store has begun a new ad campaign on local television. Before the campaign, the long term mean daily sales were $24,819. A random sample of 40 days during the new ad campaign gave a sample mean daily sale of $25,910. Does this indicate that the population mean daily sales is now more than $24,819? Use a 1% level of significance. Assume o = $1,917.

Answer 1

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score. As a percentage, is it more or less than 1%?

Answer 2

To determine whether the population mean daily sales is now more than $24,819, we can conduct a hypothesis test using the given information and a 1% level of significance.

Let's set up our null and alternative hypotheses:

Null hypothesis (H0): The population mean daily sales is not more than $24,819.
Alternative hypothesis (H1): The population mean daily sales is more than $24,819.

We can use a one-sample t-test since we have a sample mean and the population standard deviation is given. The formula for the test statistic is:

t = (sample mean - hypothesized mean) / (population standard deviation / sqrt(sample size))

Using the given values, we can calculate the test statistic:

t = (25,910 - 24,819) / (1,917 / sqrt(40))

Calculating this expression yields:

t ≈ 4.503

Next, we need to find the critical t-value for a one-tailed test with a 1% level of significance and degrees of freedom (df) = sample size -1 = 40 -1 = 39. We can use a t-table or a statistical software to find the critical t-value.

Looking up the critical t-value in the t-table with 39 degrees of freedom and a cumulative probability of 0.99 (since we are conducting a one-tailed test), we find the critical t-value to be approximately 2.423.

Since the test statistic (t ≈ 4.503) is greater than the critical t-value (2.423), we can reject the null hypothesis.

Therefore, we have evidence to suggest that the population mean daily sales is now more than $24,819 at a 1% level of significance.