Osvaldo Apodaca

2021-12-16

A Bank wishes to determine the mean income of its credit card holders and more specifically wants to investigate whether there are any differences in the mean income between males and females. Confidence interval as a statistical inference method has been considered to respond this objective.

You are required to

1) Comment on the choice of confidence intervals as a statistical inference method for the above. Describe how you will proceed to determine the corresponding confidence intervals.

2) Explain how you would use confidence intervals as evidence suggesting that in average males and females income differ.

Use Numerical examples to illustrate as and when required.

(Word count around 250 words)

You are required to

1) Comment on the choice of confidence intervals as a statistical inference method for the above. Describe how you will proceed to determine the corresponding confidence intervals.

2) Explain how you would use confidence intervals as evidence suggesting that in average males and females income differ.

Use Numerical examples to illustrate as and when required.

(Word count around 250 words)

Tiefdruckot

Beginner2021-12-17Added 46 answers

Step 1

Since, we have to investigate whether there are any differences in the mean income between males and females we perform the t-test for two groups where the test statistics is given by,

$t=\frac{(\stackrel{\u2015}{{x}_{1}}-\stackrel{\u2015}{{x}_{2}})-({\mu}_{1}-{\mu}_{2})}{\sqrt{\frac{({n}_{1}-1){s}_{1}^{2}+({n}_{2}-1){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}(\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}})}}$

Where:

${n}_{1},\text{}{n}_{2}=$ sample size for the male female groups

${s}_{1},\text{}{s}_{2}=$ sample standard deviations for the two groups

$\stackrel{\u2015}{{x}_{1}},\text{}\stackrel{\u2015}{{x}_{2}}=$ sample means for the male and female groups.

Step 2

To compute confidence intervals we can choose the confidence levels to be 0.01, 0.05, 0.10, and 0.20.

Thus, we can compute 99%, 95%, 90%, 80% confidence intervals.

Using the above t-test statistic we can compute the confidence interval as below:

$CI=(\stackrel{\u2015}{{x}_{1}}-\stackrel{\u2015}{{x}_{2}})\pm {t}_{\frac{\alpha}{2},\text{}({n}_{1}+{n}_{2}-2)}\sqrt{\frac{({n}_{1}-1){s}_{1}^{2}+({n}_{2}-1){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}(\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}})}$

Step 3

Based on the confidence interval the following conclusions can be made:

If the calculated confidence interval does not contain zero in it, then we can conclude that the average males and females’ income differ.

If the calculated confidence interval contains zero in it, then we can conclude that the average males and females’ income do not differ.

Since, we have to investigate whether there are any differences in the mean income between males and females we perform the t-test for two groups where the test statistics is given by,

Where:

Step 2

To compute confidence intervals we can choose the confidence levels to be 0.01, 0.05, 0.10, and 0.20.

Thus, we can compute 99%, 95%, 90%, 80% confidence intervals.

Using the above t-test statistic we can compute the confidence interval as below:

Step 3

Based on the confidence interval the following conclusions can be made:

If the calculated confidence interval does not contain zero in it, then we can conclude that the average males and females’ income differ.

If the calculated confidence interval contains zero in it, then we can conclude that the average males and females’ income do not differ.

Matthew Rodriguez

Beginner2021-12-18Added 32 answers

Step 1

Suppose, random variables X and Y denote income of male and female credit card holders respectively.

Here, two different groups (male and female) are used to collect data. Further we do not know population standard deviation (or variance). So, we have to use test statistic corresponding to two sample t-test.

Corresponding test statistic is given by

$t=\frac{(\stackrel{\u2015}{x}-\stackrel{\u2015}{y})-({\mu}_{x}-{\mu}_{y})}{s\times \sqrt{\frac{1}{{n}_{x}}+\frac{1}{{n}_{y}}}}$

Here,

First sample size$n}_{x$

Second sample size$n}_{y$

Sample mean of first sample$\stackrel{\u2015}{x}$

Sample mean of second sample$\stackrel{\u2015}{y}$

Pooled sample standard deviation is given by

$s=\sqrt{\frac{({n}_{x}-1)\times {s}_{x}^{2}+({n}_{y}-1)\times {s}_{y}^{2}}{{n}_{x}+{n}_{y}-2}}$

Degrees of freedom$v={n}_{x}+{n}_{y}-2$

For$100(1-\alpha )\mathrm{\%}$ confidence interval we have,

$P(-{t}_{\frac{\alpha}{2},\text{}v}t{t}_{\frac{\alpha}{2},\text{}v})=1-\alpha$ Using R-code $\sqrt{1-\frac{1-\alpha}{2},v}$

$\Rightarrow P(-{t}_{\frac{\alpha}{2},\text{}v}\frac{(\stackrel{\u2015}{x}-\stackrel{\u2015}{y})-({\mu}_{x}-{\mu}_{y})}{s\times \sqrt{\frac{1}{{n}_{x}}+\frac{1}{{n}_{y}}}}{t}_{\frac{\alpha}{2},\text{}v})=1-\alpha$

$\Rightarrow P(-s{t}_{\frac{\alpha}{2},\text{}v}\sqrt{\frac{1}{{n}_{x}}+\frac{1}{{n}_{y}}}(\stackrel{\u2015}{x}-\stackrel{\u2015}{y})-({\mu}_{x}-{\mu}_{y})s{t}_{\frac{\alpha}{2},\text{}v}\sqrt{\frac{1}{{n}_{x}}+\frac{1}{{n}_{y}}})=1-\alpha$

Suppose, random variables X and Y denote income of male and female credit card holders respectively.

Here, two different groups (male and female) are used to collect data. Further we do not know population standard deviation (or variance). So, we have to use test statistic corresponding to two sample t-test.

Corresponding test statistic is given by

Here,

First sample size

Second sample size

Sample mean of first sample

Sample mean of second sample

Pooled sample standard deviation is given by

Degrees of freedom

For

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