Question

Suppose Motorola wishes to estimate the mean talk time for its V505 camera phone before the battery must be recharged. In a random sample of 35 phones, the sample mean talk time was 325 minutes.

(a) Why can we say that the sampling distribution of x̄ is approximately normal?

(b) Construct a 94% confidence interval for the mean talk time for all Motorola V505 camera phones, assuming that σ = 31 minutes. Interpret this interval.

(c) Construct a 98% confidence interval for the mean talk time for all Motorola V505 camera phones, assuming that σ = 31 minutes. Interpret this interval.

(d) How many phones would Motorola need to test to estimate the mean talk time for all V505 camera phones within 5 minutes with 95% confidence?

Show work. You can use technology.

Answer 1

a)

The central limit theorem states that the sampling distribution of the mean of any independent, random variable will be normal or nearly normal, if the sample size is large enough.

b)
sample mean, xbar = 325
sample standard deviation, σ = 31
sample size, n = 35

Given CI level is 94%, hence α = 1 - 0.94 = 0.0600000000000001
α/2 = 0.0600000000000001/2 = 0.03, Zc = Z(α/2) = 1.88

ME = zc * σ/sqrt(n)
ME = 1.88 * 31/sqrt(35)
ME = 9.85

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (325 - 1.88 * 31/sqrt(35) , 325 + 1.88 * 31/sqrt(35))
CI = (315.1489 , 334.8511)
Therefore, based on the data provided, the 94% confidence interval for the population mean is 315.1489 < μ < 334.8511 which indicates that we are 94% confident that the true population mean μ is contained by the interval (315.1489 , 334.8511)

c)
sample mean, xbar = 325
sample standard deviation, σ = 31
sample size, n = 35

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

ME = zc * σ/sqrt(n)
ME = 2.33 * 31/sqrt(35)
ME = 12.21

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (325 - 2.33 * 31/sqrt(35) , 325 + 2.33 * 31/sqrt(35))
CI = (312.7909 , 337.2091)
Therefore, based on the data provided, the 98% confidence interval for the population mean is 312.7909 < μ < 337.2091 which indicates that we are 98% confident that the true population mean μ is contained by the interval (312.7909 , 337.2091)

d)

The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 5, σ = 31

The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.96 * 31/5)^2
n = 147.67

Therefore, the sample size needed to satisfy the condition n >= 147.67 and it must be an integer number, we conclude that the minimum required sample size is n = 148
Ans : Sample size, n = 148