Statistics - Desklib .

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STATISTICS
STUDENT NAME/ID
[Pick the date]
Question 2
Total number of student those took biology exam n1= 144,796
Total number of female student those took biology exam x1=84,199
Total number of student those took AB exam n2= 211,693
Total number of female student those took AB exam x2=102,598
90% confidence interval for the difference in the proportion of female students taking biology
exam and AB exam =?
Proportion of female student those took biology exam p1= x1/n1 = 84199/144796= 0.5815
Proportion of female student those took biology exam p2= x1/n1 = 102598/211693=0.4847
The z value for 90% confidence interval = 1.645
Standard error = 0.5815(10.5815)
144796 + 0.4847(10.4847)
211693 =0.0017
Lower limit of 90% confidence interval = (0.5815-0.4847) – (1.645 * ) = 0.0941
Upper limit of 90% confidence interval= (0.5815-0.4847) + (1.645 * )= 0.0996
Hence, 90% confidence interval =[0.0941 0.0996].
Question 5
Step 1: Null and alternative hypothesis
Null hypothesis H0: p1-p2 = 0
Alternative hypothesis H1: p1-p2 > 0
Step 2: Test statistic
n1 = 18440
1
x1 = 245
p1=245/18440 =0.01328
Now,
n2=2123
x2=45
p2 =45/2123 =0.021196
Pooled proportion p = (245+45)/(18440+2123) = 0.014103
The z value
Z value = -2.92699
Step 3: The p value
The corresponding two tailed p value = 0.998
Step 4: Significance level
The given significance level (Alpha) = 1%
Step 5: Result
It is apparent from the above that p value is higher than significance level and hence, insufficient
evidence is present to reject the null hypothesis and to accept the alternative hypothesis.
Therefore, “it cannot be said that proportion of children diagnosed with ASD in Pennsylvania is
higher than the proportion of children those are diagnosed with ASD in Utah.”
Question 3
2
Claim- West fish wholesaler is more expensive than an east fish wholesaler.
Step 1: Null and alternative hypothesis
Null hypothesis H0: μ1- μ 2 = 0
Alternative hypothesis H1: μ1- μ 2 > 0
Step 2: Test statistic
Sample mean difference x(d) =22.01/8 = 2.445
Sample standard deviation difference = sqrt{(491.84 – (22.01^2)/9}/8 = 7.3994
The t stat = (2.445-0)/(7.3994/ sqrt(9)) = 0.9915
Degree of freedom = 9-1= 8
Step 3: The p value
The corresponding two tailed p value = 0.1752
Step 4: Significance level
3

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