Question

In: Statistics and Probability

1. A random sampling of 60 pitchers from the National League and 74 pitchers from the...

1. A random sampling of 60 pitchers from the National League and 74 pitchers from the American League showed that 38 National and 36 American League pitchers had E.R.A's below 3.5.

Find the test statistic that would be used to test the claim that the proportion of the NL pitchers with E.R.A. below 3.5 is higher than the proportion of the AL pitchers with similar stats.

Round your answer to three decimal places.

2. Two independent samples are randomly selected and come from populations that are normal. The sample statistics are given below:

n1 = 47 n2 = 52

1 = 24.2 2 = 18.7

s1 = 5.0 s2 = 5.6

Find the standardized test statistic t to test the hypothesis that μ1 = μ2. Round your answer to three decimal places.

Solutions

Expert Solution

Question 1)

Given,

A random sampling of 60 pitchers from the National League : n1 = 60

And 74 pitchers from the American League: n2 = 74

38 National League pitchers E.R.A's below 3.5 : y1 = 38

36 American League pitchers had E.R.A's below 3.5: y2=36

We want to find the test statistic that the claim that the proportion of the NL pitchers with E.R.A. below 3.5 is higher than the proportion of the AL pitchers with similar stats.

Proportion of National League pitchers E.R.A's below 3.5: P̂1 = 38/60 =0.633

Proportion of American League pitchers had E.R.A's below 3.5: P̂2 = 36/74 = 0.48648649

To test the proportion of two population, we need to conduct Z test for two sample proportion

Hypothesis

Null Hypothesis H0: Population proportion of National League pitchers is less than of equal to population proportion of American League pitchers: P1 ≤ P2

Alternative hypothesis H1: Population proportion of National League pitchers is greater than the population proportion of American League pitchers: p1 > p2(Claim)

Test statistic

Where,

n1: Size of sample 1,                                        n2: Size of sample 2

y1: number positive cases in sample 1,          y2: number positive cases in sample 2

p̂1: Sample 1 proportion = y1/n1,                    p̂2: Sample 2 proportion = y2/n2

p̂ : pooled sample proportion = (y1 + y2) / (n1 + n2)

p̂ = 0.5522

By putting the all the parameters available in above test statistic formula and after solving it.

Test statistic Z = 1.699

Let’s consider the level of significance = 0.05

Zc = 1.64

P value = 0.0446

Since |Z|= 1.699 > Zc = 1.64 then concluded that the null hypothesis is rejected.

And p = 0.0446 < 0.05, it is concluded that the null hypothesis is rejected.

We can say that Population proportion of National League pitchers is greater than the population proportion of American League pitchers at greater than 0.0446 level of significance.

Question 2

Given two independent samples are randomly selected and come from populations that are normal.

n1 = 47

n2 = 52

X̅1 = 24.2

X̅2= 18.7

s1 = 5.0

s2 = 5.6

We need to find the standardized test statistic t to test the hypothesis that μ1 = μ2.

Hypothesis

H0: μ1 = μ2

H1: μ1 ≠ μ2

Standardized Test Statistic

By putting all values of available parameters and after solving it we will get

Standardized Test statistic t = 5.133


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