Many grocery stores have installed self checkout systems. Below is the number of customers using the service for a sample of 15 days at the Wal-Mart in NYC,
120 108 120 114 118 91 118 92 104 104 105 112 97 118 108 117
Is it reasonable to assume that the mean number of customers using self checkout is more than 100 per day. Use the .05 significance level.
The average $ value of all sales in the self checkout on these days were
1125 1452 1325 1055 1129 1165 1325 1275 1255 1422 1244 1351 1452 1444 1332 1111
can we assume that the average sale on self checkout is less than 1250
How would your answer to both questions change if both populations are known to be normally distributed with a standard deviation of 3 for the number of customers and 25 for the $ value of sales
Find the mean and the Z score for the values in question for both situations using the formulas indicated on a later post.
P = .05 => ±2 SD
Using the values for a normal distribution in a table in the back of your statistics text, do the respective Z scores fall within ±2 SD?
I hope this helps. Thanks for asking.
To determine whether the mean number of customers using self checkout is more than 100 per day, we can use a one-sample t-test.
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The mean number of customers is not more than 100 per day (μ <= 100)
- Alternative hypothesis (Ha): The mean number of customers is more than 100 per day (μ > 100)
Step 2: Set our significance level/alpha (α) value to 0.05 (5%).
Step 3: Calculate the test statistic:
The test statistic can be calculated using the formula:
t = (x̄ - μ0) / (s / √n)
where x̄ is the sample mean, μ0 is the hypothesized population mean, s is the sample standard deviation, and n is the sample size.
In this case:
x̄ = (120 + 108 + 120 + 114 + 118 + 91 + 118 + 92 + 104 + 104 + 105 + 112 + 97 + 118 + 108 + 117) / 15 = 109.933
s = standard deviation of the sample = √[(Σ(xi - x̄)^2) / (n - 1)] = √[(Σ(xi - 109.933)^2) / (15 - 1)] ≈ 9.986
n = 15
t = (109.933 - 100) / (9.986 / √15) ≈ 1.50
Step 4: Calculate the p-value:
Using a t-distribution table or statistical software, we can find the p-value associated with the test statistic. For a one-tailed test, we compare the p-value to the significance level (α) to determine statistical significance.
The p-value is the probability of observing a sample mean of 109.933 (or more extreme) given that the null hypothesis is true. In this case, the p-value is greater than 0.05.
Step 5: Make a decision:
Since the p-value is greater than the significance level (α = 0.05), we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the mean number of customers using self checkout is more than 100 per day.
Now let's move on to the second question about average sales on self checkout.
To determine whether the average sale is less than 1250, we can also use a one-sample t-test.
Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The average sale is not less than 1250 ($μ >= 1250)
- Alternative hypothesis (Ha): The average sale is less than 1250 ($μ < 1250)
Step 2: Set our significance level/alpha (α) value to 0.05 (5%).
Step 3: Calculate the test statistic:
Using a similar approach as before, we calculate the test statistic using the formula:
t = (x̄ - μ0) / (s / √n)
In this case:
x̄ = (1125 + 1452 + 1325 + 1055 + 1129 + 1165 + 1325 + 1275 + 1255 + 1422 + 1244 + 1351 + 1452 + 1444 + 1332 + 1111) / 16 ≈ 1287.8125
s = √[(Σ(xi - x̄)^2) / (n - 1)] = √[(Σ(xi - 1287.8125)^2) / (16 - 1)] ≈ 131.838
n = 16
t = (1287.8125 - 1250) / (131.838 / √16) ≈ 1.121
Step 4: Calculate the p-value:
Using a t-distribution table or statistical software, we find the p-value associated with the test statistic. In this case, the p-value is greater than 0.05.
Step 5: Make a decision:
Since the p-value is greater than the significance level (α = 0.05), we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the average sale on self checkout is less than 1250.
If both populations were known to be normally distributed with a standard deviation of 3 for the number of customers and 25 for the $ value of sales, we could use the z-test instead of the t-test. The steps would be similar, but instead of using the t-distribution, we would use the standard normal distribution (z-distribution) to calculate the z-statistic and find the associated p-values.
However, in this scenario, we do not have information indicating that the populations are normally distributed with known standard deviations. Therefore, we need to rely on the t-test as described above.