Question

Gross sales receipts (in thousands of dollars) are recorded daily for a sample of 20 small businesses in the New York metropolitan area for a one-day period in March.  The mean of the sample is 283.2 and the standard deviation of the sample is 118.

The sample results are as follows.

521.0	299.6	430.0	325.0	170.5	248.6	125.5	91.2	353.3	268.6
359.7	207.7	392.1	336.3	467.0	300.3	150.6	139.1	263.6	214.3

a company is considered to be “successful” if its daily sales exceed $200,000.

To estimate the true percentage of “successful” small businesses, how large a sample must be taken to insure the estimate is off by no more than +3.2% with 99% certainty?

	a.	n = 1326
	b.	n = 661
	c.	n = 1247
	d.	n = 938

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 200
Alternative hypothesis: u > 200

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 26.3856
DF = n - 1

D.F = 19
t = (x - u) / SE

t = 3.15

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 3.15.

P-value = P(t > 3.15)

Use the t-value calculator for finding p-values.

P-value = 0.003

Interpret results. Since the P-value (0.003) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that

2) The required sample size is 1621.

M.E = 0.032

Confidence level = 99%

p = 0.50

n = 1620.0625

n = 1621