Question

A recent study by a marking research firm shows that the average expenditure on Valentine’s Day is $100.89. A survey was conducted to analyze the difference between spending levels between male and female consumers. The average expenditure in the sample of 40 male consumers was $135.67, and the average expenditure in a sample of 30 female consumers was $68.64. Based on past studies, the population standard deviation for male consumers is assumed $35, and that for female consumers is assumed $20. Develop a 99% confidence interval for the difference between the two population means.

Answer 1

Answer:

$67.03-2.58\sqrt{\frac{35^2}{40}+\frac{20^2}{30}}=49.92$  

$67.03+2.58\sqrt{\frac{35^2}{40}+\frac{20^2}{30}}=84.14$  

So on this case the 99% confidence interval would be given by $49.92 \leq \mu_M -\mu_F \leq 84.14$  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X_M =135.67$ represent the sample mean for male

$\bar X_F =68.64$ represent the sample mean for female

nM=40 represent the sample of amles

nF=30 represent the sample of females

$\sigma_M =35$ population standard deviation for males

$\sigma_F =20$ population standard deviation for fmale

$\mu_M -\mu_F$ parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_M}{n_1}+\frac{\sigma^2_F}{n_F}}$ (1)  

The point of estimate for $\mu_M -\mu_F$ is just given by:

$\bar X_M -\bar X_F =135.67-68.64=67.03$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$  

Confidence interval

Now we have everything in order to replace into formula (1):  

$67.03-2.58\sqrt{\frac{35^2}{40}+\frac{20^2}{30}}=49.92$  

$67.03+2.58\sqrt{\frac{35^2}{40}+\frac{20^2}{30}}=84.14$  

So on this case the 99% confidence interval would be given by $49.92 \leq \mu_M -\mu_F \leq 84.14$  

Answer 2

Answer:

Step-by-step explanation: