The Lawn Service announces that it will maintain an average cost per customer of $35 per hour. Assume that costs are distributed normally. A random sample of 18 clients showed that the average cost is $32.50 per hour with a population standard deviation of $8.10. Does the data support the claim that the average hourly cost is less than $35.00? Use a significance level of 5%. 1. The alternate hypothesis is: Ha: μ=35 Ha: μ>35 Ha: μ<32.5 Ha: μ<35 2. The test statistic is: -1.65 -1.31 -2.5 1.65 3. For a significance level of 5% the null hypothesis could be rejected?: a.Noo, because the test statistic is less than the z in the table b. No, because the test statistic is greater than the z in the table c. Yes, because the test table is greater than the z in the table d.Yes, because the teststat is less than the z in the table
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answer:
The null hypothesis, denoted as H0, is a statement that assumes there is no significant difference or relationship between two variables. In this case, the null hypothesis is that the average hourly cost is $35.00.
The alternate hypothesis, denoted as Ha, is a statement that contradicts the null hypothesis and suggests that there is a significant difference or relationship between two variables. In this case, the alternate hypothesis is that the average hourly cost is less than $35.00.
To determine if the data supports the claim that the average hourly cost is less than $35.00, we need to perform a hypothesis test. The test statistic used for this type of hypothesis test is the z-score.
The formula for calculating the z-score is:
z = (x - μ) / (σ / √n)
Where:
x is the sample mean (in this case, $32.50)
μ is the population mean (in this case, $35.00)
σ is the population standard deviation (in this case, $8.10)
n is the sample size (in this case, 18)
Calculating the z-score:
z = (32.50 - 35.00) / (8.10 / √18)
z = -1.50 / (8.10 / 4.2426)
z ≈ -1.50 / 1.9109
z ≈ -0.7851
The test statistic is approximately -0.7851.
Now, we need to compare the test statistic to the critical value in the z-table. Since we are testing if the average hourly cost is less than $35.00, we are performing a one-tailed test with a significance level of 5%.
Looking up the critical value in the z-table, we find that the z-value corresponding to a significance level of 5% is approximately -1.645.
Since the test statistic (-0.7851) is greater than the critical value (-1.645), we fail to reject the null hypothesis. This means that the data does not support the claim that the average hourly cost is less than $35.00 at a significance level of 5%.
To summarize:
1. The alternate hypothesis is: Ha: μ>35
2. The test statistic is: -0.7851
3. For a significance level of 5%, the null hypothesis could be rejected?: b. No, because the test statistic is greater than the z in the table.
Alli <3
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