A machine that produces ball bearings reports that the average diameter is not 0.5 cm. A sample of 10 ball bearings was measured and it was found that the average diameter is 0.48 cm, and the population standard deviation is 0.05 cm. (a = 0.05)
a) State the null hypothesis and the alternate hypothesis. (2 Marks)
b) Find the critical values. (1 Mark )
c) Calculate the test statistic (2 Marks)
d) State your decision regarding the null hypothesis (1 Mark )
e) Summarize the results. (1 Mark )
f) If the average diameter is less than 0.5 cm and a = 0 01,. what is the hypothesis and the critical values (3 Marks)
a) Null hypothesis and the alternate hypothesis
"H_0:\\mu=0.5"
"H_a:\\mu\\neq0.5"
b) Critical values
Level of significance is 0.05
"z_\\frac{\\alpha}{2} =z_\\frac{0.05}{2}=1.96"
c) Calculate the test statistic
"z=\\frac{x-\\mu}{\\frac{\\sigma}{\\sqrt{n}}}"
"z=\\frac{0.48-0.5}{\\frac{0.05}{\\sqrt{10}}}=-1.26"
d) Decision regarding the null hypothesis
Hence, test statistic is less than the critical value, so reject null hypothesis.
e) Summarize the results
It can conclude that the average diameter of ball bearings is not equal to 0.5 cm.
f) Revised hypothesis and the critical values
If the average diameter is less than 0.5 cm then null and alternative hypothesis are:
"H_0:\\mu=0.5"
"H_a:\\mu>0.5"
If alpha is 0.01 then the critical value is:
"z_\\frac{\\alpha}{2} =z_\\frac{0.01}{2}=2.575"
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