Question

The Port Authority sells a wide variety of cables and adapters for electronic equipment online. Last year the mean value of orders placed with the Port Authority was $47.28, and management wants to assess whether the mean value of orders placed to date this year is the same as last year. The values of a sample of 49,896 orders placed this year are collected and recorded in the file PortAuthority.
A) At the 5% level of significance, test whether the mean value of this year’s orders differs from last year’s mean orders.
B) Port Authority also wants to determine if the gender profile of its customers has changed from last year when 65% of orders were placed by male customers. Test at 5% level.
C) At the 5% level of significance, can we conclude that the mean value of orders by male customers is higherthan the mean value of orders made by female customers?

Answer 1

Answer:

Step-by-step explanation:

Hello!

Data set in attachment. (100 orders)

a)

The variable of interest is

X: value of the orders placed for cables and adapters for electronic equipment over the course of the year.

Using the mean value of the orders placed the year before as the population mean μ= $47.28

The sample size is n= 100

Sample mean of the value of orders $\frac{}{X} = \frac{4544.69}{100} = 45.45$

Sample standard deviation: $S= \sqrt{(\frac{1}{n-1} )[sumX^2-\frac{(sumX)^2}{n} ]} = \sqrt{\frac{1}{99}[239000.55-\frac{(4544.69)}{100} ] } = 18.11$

The interest is to test if there is any difference in the mean value of the orders regarding last year, so the hypotheses are:

H₀: μ = $47.28

H₁: μ ≠ $47.28

α: 0.05

Applying the Central Limit Theorem, since the sample size is big enough, the statistic to use is an approximation to of the standard normal distribution.

$Z= \frac{\frac{}{X} - Mu}{\frac{S}{\sqrt{n} } }$ ≈ N(0;1)

$Z_{H_0}= \frac{45.45-47.28}{\frac{18.11}{\sqrt{100} } } = -1.01$

p-value: 0.312495

To decide using the p-value approach you have to follow the decision rule:

p-value ≤ α, reject the null hypothesis.

p-value > α, do not reject the null hypothesis.

In this case the p-value is greater than the significance level, so the decision is to not reject the null hypothesis.

So at 5% significance level you can conclude that the mean value of orders placed the current year is equal to the mean value of orders placed last year.

b)

For this item the parameter of interest is the proportion of males that placed an order: p

The port Authority wishes to test if it is different from last years proportion of p=0.65

The hypotheses are:

H₀: p = 0.65

H₁: p ≠ 0.65

α: 0.05

Sa sample proportion for this item is the number of males that placed an order in the sample

p'= 62/100= 0.62

$Z_{H_0}= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } } = \frac{0.62-0.65}{\sqrt{\frac{0.65*0.35}{100} } } = -0.628= -0.63$

p-value: 0.528695

p-value ≤ α, reject the null hypothesis.

p-value > α, do not reject the null hypothesis.

In this case the p-value is greater than the significance level, so the decision is to not reject the null hypothesis.

So at 5% significance level you can conclude that the proportion of orders placed by males is equal yo 65%.

C)

We have only one sample, where the orders were classified regarding the gender of the person that placed the order, so:

The variable of interest is

X: number of orders placed by males

X~Bi(n$_x$;p$_x$)

The complementary variable can be defined as:

Y: number of orders placed by females.

Y~Bi(n$_Y$;p$_Y$)

Where p$_Y$= 1 - p$_x$

Worst case scenario both probabilities are equal p$_Y$= p$_x$= 0.5, so to test if the proportion of males in greater than the proportion of females who placed orders, we can say that p$_x$ > 0.5

The hypotheses are:

H₀: p$_x$ ≤ 0.5

H₁: p$_x$ > 0.5

p$_x$ ≤ 0.5 ⇒ If the proportion of orders placed by men is less than 0.5, then we can conclude that the proportion of orders placed by females is greater than the proportion of orders placed by males.

p$_x$ > 0.5 ⇒ If the proportion of orders placed by men is greater than 0.5, then we can conclude that the proportion of orders placed by males is greater than the proportion of orders placed by females.

$Z_{H_0}= \frac{p'_x-p_x}{\sqrt{\frac{p_x*(1-p_x)}{n} } } = \frac{0.62-0.65}{\sqrt{\frac{0.5*0.5}{100} } } = -0.6$

p-value: 0.274253

p-value ≤ α, reject the null hypothesis.

p-value > α, do not reject the null hypothesis.

In this case the p-value is greater than the significance level, so the decision is to not reject the null hypothesis.

At 5% significance level you can conclude that the proportion of orders placed by males is no greater than the proportion of orders placed by females.