A production line is designed to fill bottles with 200 ounces of laundry detergent. A manager wants to test the line's output to see if the machines really are filling the bottles with 200 ounces, or if they contain some other amount on average. This is a test to see whether the population mean is 200 ounces or some other value. To do this, a sample of 25 bottles is taken. The sample mean is 198 ounces, the sample standard deviation s is 3 ounces. The population standard deviation σ is unknown.
1) Is this test for the hypothesis about the population mean an upper-tail, lower-tail, or two-tailed test?
Lower-tail |
|
Upper-tail |
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Two-tailed |
2) Which of the following is the null hypothesis about the population mean?
The average amount of detergent in the bottles is some amount other than 200 ounces. |
|
The average amount of detergent in the bottles really is 200 ounces. |
3) Which distribution must be used to test this hypothesis about the population mean?
F distribution |
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Student's t-distribution |
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χ2 (chi-square) distribution |
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Standard normal distribution |
4) What are the degrees of freedom for the test about the population mean? (Integer, no rounding, no decimals)
5) Compute the necessary test statistic to test the hypothesis about the population mean. Round your answer to three decimal places and be sure to include the minus sign if negative.
6) Suppose that the hypothesis about the population mean is at the α = 0.01 (1%) significance level. What is the critical value of the test statistic? Three decimal places. You do not need a +/- sign.
7) Should the null hypothesis about the mean be rejected?
Solution :
= 200
= 198
s = 3
n = 25
1) This is the two tailed test .
The null and alternative hypothesis is
H0 : = 200
Ha : 200
2)The average amount of detergent in the bottles really is 200 ounces.
3) Student's t-distribution
4) Degrees of freedom = n-1 = 25-1 = 24
5) Test statistic = t
= ( - ) / s / n
= (198 -200) / 3/ 25
= -3.333
P (t< -3.333) = 0.0028
P-value = 0.0028
6) = 0.01
Ths two tailed test critical value is = 2.797
0.0028 < 0.01
7) Reject the null hypothesis .
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