The heights (in cm) for a random sample of 60 males were measured. The sample me
ID: 3246788 • Letter: T
Question
The heights (in cm) for a random sample of 60 males were measured. The sample mean is 166.55, the standard deviation is 12.57, the sample kurtosis is 0.12, and the sample skewness is-0.23. The following table shows the heights subdivided into non-overlapping intervals. For the goodness-of-fit test for normality, the null and alternative hypothesis are ______. H_0: Heights do not follow a normal distribution with mean 166.55 and standard deviation 12.46, H_A: Heights follow a normal distribution with mean 166.55 and standard deviation 12.46 H_0: Heights follow a normal distribution with mean 166.55 and standard deviation 12.57, H_A: Heights do not follow a normal distribution with mean 166.55 and standard deviation 12.57: H_0: Heights do not follow a normal distribution with mean 166.55 and standard deviation 12.57, H_A: Heights follow a normal distribution with mean 66.55 and standard deviation 2 5 H_0: Heights follow a normal distribution with mean 166.55 and standard deviation 12.46, H_A: Heights do not follow a normal distribution with mean 166.55 and standard deviation 12.46Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Height follows a normal distribution with mean 166.55 and standard deviation 12.57.
Alternative hypothesis: Height do not follow a normal distribution with mean 166.55 and standard deviation 12.57.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.
Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = k - 1 = 5 - 1
D.F = 4
(Ei) = n * pi
2 = [ (Oi - Ei)2 / Ei ]
2 = 7.70
where DF is the degrees of freedom.
The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 7.70.
We use the Chi-Square Distribution Calculator to find P(2 > 7.70) = 0.10
Interpret results. Since the P-value (0.10) is more than the significance level (0.05), we have to accept the null hypothesis.
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