STAT 200 Week 6 Homework Problems 9.1.2 Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology exam 84,199 of them were female....

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Answered Same DayJul 26, 2021

Answer To: STAT 200 Week 6 Homework Problems 9.1.2 Many high school students take the AP tests in different...

Atreye answered on Jul 27 2021
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Solution 9.1.2
Let be the proportion of female who took the biology exam.
Let be the proportion of female who took the calculus AB exam.
and are calculated as below:
The confidence interval for the difference of two proportions is calculated using the formula:

For the 90% co
nfidence interval the value of level of significance,  and . With 0.05 level of significance, the z value is obtained as 1.64 using the standard normal distribution table.
The 90% confidence interval is calculated as below:
There is 90% certainty that the difference between the two proportions lies within (0.094, 0.0996).
Solution 9.1.2
Let be the proportion of eight years old diagnosed with ASD in the state of Pennsylvania
Let be the proportion of eight years old diagnosed with ASD in the state of Utah.
and are calculated as below:
The appropriate null and alternative hypothesis is stated as below:
The sample pooled proportion is calculated as below:
The test statistic is calculated as below:
With 0.05 level of significance, the z value is obtained as 1.2816 using the standard normal distribution table.
Conclusion: The null hypothesis is not rejected. It cannot be concluded that proportion of children diagnosed with ASD in Pennsylvania is more than the proportion of children diagnosed with ASD in Utah.
Solution 9.2.3
Paired t-test will be appropriate here to use.
The null and alternative hypothesis is:
The calculation of test statistic is given below:
Degrees of freedom for t=0.9915 is 9 -1=8.
The p-value for t=0.9915 is 0.8245.
Conclusion:
Since, the p-value is larger than the level of significance, the null hypothesis is failed to reject and it cannot be concluded that the west coast fish wholesalers are more expensive than east coast wholesalers.
Solution 9.2.6
To estimate the mean difference for confidence interval, the sample statistic will be d=value of 6th –value of 13th.
The confidence interval is 90%, so the level of significance will be
The degrees of freedom are 10 -1=9
Using a t-distribution table, the t-score for α = 0.05 and df = 9 is 1.833.
The sample mean, sample standard deviation and standard error are calculated as below:
The interval is calculated as below:
Therefore, the estimated mean difference in traffic count between 6th and 13th using 90% level of confidence lies within (-799586.3, 803257.9).
Solution 9.3.1:
The appropriate null and alternative hypothesis is stated as below:
For males:
For Females:
The degrees of freedom are calculated as below:
The pooled standard deviation is calculated as below:
The t statistic is calculated as below:
The corresponding p-value is 0.000.
Conclusion:
Since p-value is less than 0.01 so we reject the null hypothesis.
The result is statistically significant at .01 level of significance. Sufficient evidence exists to
support the claim that the mean...
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