# Statistics - Desklib .

Added on 2023-05-29

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STATISTICS

STUDENT NAME/ID

[Pick the date]

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(1−0.5815)

144796 + 0.4847(1−0.4847)

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

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(1−0.5815)

144796 + 0.4847(1−0.4847)

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

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

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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