Question

1. A random sample of 82 customers, who visited a department store, spent an average of $71 at this store. Suppose the standard deviation of expenditures at this store is O = $19. What is the e 98% confidence interval for the population mean? 2. A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32 Assuming that the population has a normal distribution, what is the 90% confidence interval for the population mean? 3. A researcher wants to estimate the mean age of all Business Week readers at a 99% confidence level. She wants the margin of error to be within 3.1 years of the population mean. The standard deviation of ages of all Business Week readers is 10.04 years. What is the sample size that will yield a margin of error within 3.1 of the population mean? 4. The Labor Bureau wants to estimate, at a 90% confidence level, the proportion of all households that receive welfare. A preliminary sample showed that 18.5% of households in this sample receive welfare. What is the he sample size that would limit the margin of error to be within 0.036 of the population proportion? We are using the mean of a sample as a point estimate for the mean of a normal distribution with a standard deviation of 5. The margin of error, with 95% confidence, for this estimate is 0.860. What is the sample size?

Answer 1

(1) $\bar{x}$ = 71, $\sigma$ = 19, n = 82

The Critical value at $\alpha$ = 0.02 is 2.33,

The Lower Limit = $\bar{x}$ - Zcritical * $\sigma$ / sqrt(n) = 71 - 2.33 * 19 / sqrt(82) = 66.12

The Lower Limit = $\bar{x}$ - Zcritical * $\sigma$ / sqrt(n) = 71 - 2.33 * 19 / sqrt(82) = 75.88

The 98% CI is (66.12, 75.88)

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(2) $\bar{x}$ = 123.4, s = 18.32, n = 25

The Critical value at $\alpha$ = 0.10, df = n - 1 = 24 is 1.71,

The Lower Limit = $\bar{x}$ - tcritical * s / sqrt(n) = 123.4 - 1.71 * 18.32 / sqrt(25) = 117.13

The Lower Limit = $\bar{x}$ - tcritical * s / sqrt(n) = 123.4 - 1.71 * 18.32 / sqrt(25) = 129.67

The 90% CI is (117.13, 129.67)

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(3) Given ME = 3.1, $\sigma$ = 10.04, $\alpha$ = 0.01

The Zcritical at $\alpha$ = 0.01 is 2.58

The ME is given by :

$ME = Zcritical * \frac{\sigma}{\sqrt{n}}$

Squaring both sides we get: (ME)² = (Z critical)² * $\sigma$ ² / n

Therefore n = (Zcritical * $\sigma$ / ME)² = (2.58 * 10.04 / 3.1)² = 69.82

Therefore n = 70 (Taking it to the next whole number)

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(4) The ME = Zcritical * SQRT(p * q/n).

Squaring and solving, n = (Zc/ME)² * p * q

Here ME = 0.036, p = 18.5% = 0.185, q = 1 - p = 0.815

The Zc at $\alpha$ = 0.1 is 1.65

Therefore n = (1.65 / 0.036)² * 0.185 * 0.815 = 316.73

Therefore n = 317 (Rounding to the nearest Integer)

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(5) Given ME = 0.86, $\sigma$ = 5, $\alpha$ = 0.05

The Zcritical at $\alpha$ = 0.01 is 1.96

The ME is given by :

$ME = Zcritical * \frac{\sigma}{\sqrt{n}}$

Squaring both sides we get: (ME)² = (Z critical)² * $\sigma$ ² / n

Therefore n = (Zcritical * $\sigma$ / ME)² = (1.96 * 5 / 0.86)² = 129.85

Therefore n = 130 (Taking it to the next whole number)

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