Statistics Null and alternative hypothesis

Question

Historically, the mean dollar spent for in-store retail purchases (charges) on the Vista
Charge Card the day after Thanksgiving Day(BLACK FRIDAY) for customers who use their
Vista Charge Card that day for in-store retail shopping is $870 . The population standard deviation is not known.
Because of increased use of on-line purchasing, Vista’s Vice
President of Electronic Marketing believes in-store retail purchasing has decreased and
randomly samples 500 customer accounts, and found that those customer’s spent a
mean dollar amount of 855 dollars for in-store retail purchases withs = 60 dollars

a)State the null and alternative hypothesis in symbols and in words.
b)Calculate the expected results for the hypothesis test sampling distribution, assuming
the null hypothesis is true. (i.e., name and graph of sampling distribution, mean and standard error)
c)Identify the standard distribution that best approximates the sampling distribution.
d)Formulate the decision rule(use can use either critical test scores or p-values)
e)Determine the statistical sample results and the test statistic.
f)Determine the conclusion in terms of the null and alternative hypotheses. Does the sample results indicate that the vice president’s claim is supported at a=5%?

Answer 1

Given Information:

1. Historical (population) mean: μ₀ = $870
2. Sample mean: x̄ = $855
3. Sample standard deviation: s = $60
4. Sample size: n = 500
5. Significance level: α = 0.05

Vista’s historical average for in-store retail purchases on Black Friday is $870. A new sample of 500 customer accounts showed an average spending of $855. The sample standard deviation was $60. The Vice President of Electronic Marketing believes that in-store spending has gone down, possibly due to the rise in online shopping. We are going to test whether this sample provides enough evidence to support that belief.

To begin, we set up our hypotheses. The null hypothesis is that the average spending has stayed the same, so the population mean is still 870 dollars. This is written as H₀: μ = 870. The alternative hypothesis is that the average has decreased, so H₁: μ < 870. This is a one-tailed test because we are specifically looking for evidence of a decrease, not just any change.

Next, we assume the null hypothesis is true and describe what the sampling distribution should look like. Since we are using a large sample size of 500 and we do not know the population standard deviation, we technically should use a t-distribution, but with such a large n, the t-distribution is almost identical to the normal distribution, so we use the normal distribution to make things easier.

The mean of the sampling distribution (if H₀ is true) is still 870. To calculate the standard error of the mean, we divide the sample standard deviation by the square root of the sample size. So, SE = 60 / √500, which is approximately 60 / 22.36, giving us a standard error of about 2.683. This means if the true average spending is still 870 dollars, the sample means from random samples of size 500 should mostly fall within about 2.683 dollars of 870, just from natural variation.

Now that we know the expected distribution under the null, we can figure out how unusual our sample mean of 855 is. To do that, we calculate a z-score using the formula:

z = (x̄ - μ) / SE = (855 - 870) / 2.683 = -15 / 2.683 ≈ -5.59

This tells us that our sample mean is 5.59 standard errors below the expected mean. That’s very far in the left tail of the normal distribution. To decide whether that is far enough to reject the null, we compare it to the critical value for a left-tailed test at the 5% significance level. The critical z-score for 5% in the lower tail is about -1.645. Since -5.59 is much smaller than -1.645, it falls in the rejection region.



Therefore, we reject the null hypothesis. This means our sample provides strong statistical evidence that average in-store spending on Black Friday has decreased. So yes, based on this test, the Vice President’s belief is supported at the 5% significance level.