Mathematics, 08.03.2021 19:40, malaysiae6321

Eddie Clauer sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. Random samples of sales receipts were studied for mail-order sales and internet sales, with the total purchase being recorded for each sale. A random sample of 8 sales receipts for mail-order sales results in a mean sale amount of $69.30 with a standard deviation of $26.25. A random sample of 17 sales receipts for internet sales results in a mean sale amount of $75.90 with a standard deviation of $28.25. Using this data, find the 90% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases. Assume that the population variances are not equal and that the two populations are normally distributed. 1. Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
2. Find the Staandard error of the sampling distrbution to be used in constructing the confidence interval

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answered: KennyMckormic

1. The critical value to be used for constructing the confidence interval is 1.415

2. The standard error of the sampling distribution to be used in constructing the confidence interval, is approximately 6.260

Step-by-step explanation:

The number of mail-order receipts, n₁ = 8 sales receipts

The mean amount of sale, $\bar x_1$ = $69.30

The standard deviation of the sample, σ₁ = $26.25

The number of internet sales receipts, n₂ = 17 sales receipts

The mean amount of sale, $\bar x_2$ = $75.90

The standard deviation of the sample, σ₂ = $28.25

1. Where the population variance are not equal, we have;

$\left (\bar{x}_1-\bar{x}_{2} \right ) - t_{c}\sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + t_{c}\sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}$

Where;

t_c = The critical value for constructing the confidence interval

$\sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}$ = $S.E_{\bar{x}_1-\bar{x}_{2}} =$ The standard error for constructing the confidence interval

The degrees of freedom, df = n₁ - 1 = 8 - 1 = 7

The t-value at 90% confidence level, and degrees of freedom of 7, $t_{90\%, 7}$ , from the tables is given as follows;

t_c = $t_{90\%, \, 7}$ = 1.415

The critical value for constructing the confidence interval = t_c = 1.415

2. The standard error of the sampling distribution to be used in constructing the confidence interval, $S.E._{\bar{x}_1-\bar{x}_{2}}$ , is given as follows

$S.E_{\bar{x}_1-\bar{x}_{2}} = \sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}$

$S.E._{\bar{x}_1-\bar{x}_{2}} = \sqrt{\dfrac{26.25^{2}}{8}-\dfrac{28.25^{2}}{17}} \approx 6.260$

$S.E._{\bar{x}_1-\bar{x}_{2}}$ ≈ 6.260

The standard error of the sampling distribution to be used in constructing the confidence interval, $S.E._{\bar{x}_1-\bar{x}_{2}}$ ≈ 6.260

answered: Guest

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

answered: Guest

the car was traveling 40 mph before it started skidding. ( 2 points ) 17. was the driver were her brakes working at 75% efficiency or better? ( 2 points ) yes 29.

step-by-step explanation: