Question 1201978: When 40 people use the Weight Watchers diet for one year, their mean weight loss was 3.0 pounds. Historically the standard deviation is s = 4.9 pounds. Use a 0.01 significance level to test the claim that the mean weight loss is greater than 0.
a) State the null and alternate hypotheses.
b) Explain whether your test will be left, right or two-tailed.
c) State the test statistic and p-value of your test. Round your answers to four-decimal places.
d) Explain your conclusion to reject or fail to reject the null hypothesis in terms of p and s.
e) Interpret your conclusion in the context of this problem.
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Thank you!
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
Hypotheses:
Null: mu = 0
Alternate: mu > 0
The equal sign is ALWAYS in the null.
This is to lock down the parameter we're trying to test.
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Part (b)
We have a right-tailed test because of the "greater than" sign in the alternate hypothesis.
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Part (c)
Your teacher should have used sigma in place of "s".
xbar = 3.0 = sample mean
mu = 0 = hypothesized population mean weight loss
sigma = 4.9 = population standard deviation
n = 40 = sample size
Because sigma is known, we use the standard normal Z distribution.
Test statistic:
z = (xbar - mu)/(sigma/sqrt(n))
z = (3.0 - 0)/(4.9/sqrt(40))
z = 3.87 approximately
Now use a table such as this one
https://www.ztable.net/
Similar tables are found in the back of your stats textbook.
According to such a table,
P(Z < 3.87) = 0.99995 approximately
So,
P(Z > 3.87) = 1-P(Z < 3.87)
P(Z > 3.87) = 1-0.99995
P(Z > 3.87) = 0.00005
This is the approximate p-value.
It's the approximate area under the standard normal Z curve to the right of z = 3.87
When rounding to four decimal places we go from 0.00005 to 0.0001
Summary:
test statistic: z = 3.87
p-value = 0.0001
Each value is approximate.
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Part (d)
p value = 0.0001 approximately
alpha = 0.01
Since the p-value is smaller than alpha, we reject the null.
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Part (e)
Rejecting the null means we side with mu > 0 in the alternate hypothesis.
We conclude that the average weight loss is greater than 0 pounds.
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