HURRY PLEASE I NEED HELP
1 answer:
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Answer:
Step-by-step explanation:
The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.
To find the percentile for the television weights, use the formula:
, where
is the average of the set,
is some constant relevant to the percentile you're finding, and
is one standard deviation.
As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute ,
, and
:
Therefore, the 90th percentile weight is 5.1282 pounds.
Repeat the process for calculating the 10th percentile weight:
The difference between these two weights is .
There is a missing content in the question.
After the statements and before the the options given; there is an omitted content which says:
Referring to Table 16-5, in testing the coefficient of X in the regression equation (0.117) the results were a t-statistic of 9.08 and an associated p-value of 0.0000. Which of the following is the best interpretation of this result?
Answer:
C. The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05).
Step-by-step explanation:
From the given question:
The resulting regression equation can be represented as:
where;
the estimated number of contracts in a quarter X is the coded quarterly value with X = 0
the first quarter of 2010 Q1 is a dummy variable equal to 1 in the first quarter of a year and 0 otherwise
Q2 is a dummy variable equal to 1 in the second quarter of a year and 0 otherwise
Q3 is a dummy variable equal to 1 in the third quarter of a year and 0 otherwise
Our null and alternative hypothesis can be stated as;
Null hypothesis :
The quarterly growth rate in the number of contracts is not significantly different from 0% (? = 0.05)
The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05)
The decision rule is to reject the null hypothesis if the p-value is less than 0.05.
From the missing omitted part we added above; we can see that the t-statistics value = 9.08 and the p-value = 0.000 .
Conclusion:
Thus; we reject the null hypothesis and accept the alternative hypothesis. i.e
The quarterly growth rate in the number of contracts is significantly different from 0% (? = 0.05)