Download Question 2 - JustAnswer

Question 2:
Nominal information is basically non-numerical. It is just categories that have no
number attached or numerical relationship. A work related survey question could be
simply “What is your gender” with possible responses of male or female.
Question 3:
The null hypothesis is that the mean ounces of root beer bottled is 32. The
alternative hypothesis is that the mean is not 32 ounces. Use the t-satistic to test the
hypothesis:
t = (32.082 – 32) / (s / SQRT(n)) = .082 / (.0995 / 3) = .082 / .0332 = 2.47
where 32.082 is the sample mean, 32 is the hypothesised mean, s is the sample standard
deviation, n is the sample size, and SQRT is the square root. This t statistic has 8 (n-1)
degrees of freedom. Using either a table of the t-distribution or a t distribution calculator
we find the probability P(|t| > 2.47) = .039. |t| is the absolute value since this is a twosided test. Using alpha = .05 we can reject the null hypothesis in favor of the alternative
hypothesis. The conclusion is that the means ounces of soda in the bottle is not equal to
32 but greater than 32 since the t-statistic is positive.
Question 4:
Z = (980 – 1000) / [50 / SQRT(36)] = -20 / [50 / 6] = -20 / 8.333 = - 2.4.
The underlying difference between the T-statistic and the Z-statistic is that in the latter
you assume you know the standard deviation instead of having to estimate it from the
sample. This points out the inconsistency in the question where a Z-statistic is asked for
but a sample standard deviation is given.
Question 5:
The average running time of broadway shows is 2 hours 12 minutes = 132
minutes. Again, inconsistency occurs in this question by asking for a Z-statistic while
giving sample estimates.
Z = (125 – 132) / [11 / SQRT(8)] = -7 / [11 / 2.828] = -7 / 3.89 = - 1.799.
At the 95% confidence level (alpha = .05), |Z| > 1.96 would be required to reject the null
hypothesis that the mean time is 132 minutes. So, the conclusion is that the producer is
correct.
Question 6:
The null hypothesis, H1, is that there is no difference between the strength of
policemen and firemen. The alternative hypothesis, Ha, is that policemen are stronger.
Question 7:
If t is from a one sample test with sample size = 22 then the degrees of freedom
are 21. From a t distribution table or calculator, the critical t value for an alpha = .05
(95% level of confidence) is 2.08. If t is from a two sample test, then the degrees of
freedom are 20 and the critical t value is 2.086.
Question 8:
t = (43,228 – 40,000) / [4000 / SQRT(20)] = 3228 / 894.4 = 3.609. From a t
distribution table or calculator, the critical t value (19 degrees of freedom) for an alpha =
.05 (95% level of confidence) is 2.093. The calculated t, 3.609, is greater than 2.093 so
we can reject the null hypothesis of no difference in the sample mean and the
hypothesized mean and conclude that the graduates in the actuarial field earn more than
$40,000.
Question 9:
The answer to this question depends on exactly what is meant by “variable”. Z
has either 3 or 4 quantities. It has three for a hypothesized value (not a mean). They are
the hypothesized value, true mean, and standard deviation. In addition, if the
hypothesized value is a mean it also has the sample size n. The T has the four quantities
of sample mean, hypothesized mean, standard deviation, and sample size.
All quantities are in an interval scale.
Question 10:
The 95% level of significance critical value for a Z statistic is 1.96 found in either
a standard normal distribution table or by a Z calculator.