Solution of Class test –II

  1. Find z so that, (i) \(P(Z<-z)=0.025\) and (ii) \(P(Z>z)=0.005\).
    Solution

(i) \(P(Z<-z)=0.025 \implies \Phi (-z)=0.025 \implies -z=\Phi^{-1}(0.025)=-1.96\)

Hence, \(z=1.96\)

(ii) \(P(Z>z)=0.005 \implies P(Z\le z)=1-0.005=0.995 \implies z=\Phi^{-1}(0.995)=2.58\)

  1. Suppose, we draw a sample of size \(n\) which is sufficiently large from population with mean \(\mu\) and standard deviation \(\sigma\). What is the sampling distribution of \(\bar x\)?

Solution Since, sample size \(n\) is sufficiently large, so, according to central limit theorem (CLT) sampling distribution of \(\bar x\) will be approximately normal that is \(\bar x \sim N(\mu,\sigma/\sqrt n)\) (approx).

  1. In 36 randomly selected seawater samples, the mean sodium chloride concentration was 23 cubic centimeters per cubic meter. Assume the population standard deviation is 6.7 cubic centimeters per cubic meter. Construct the 95% confidence intervals for the population mean. Interpret the result.

Solution

Here, \(n=36, \bar x=23 , \sigma= 6.7\) \(1-\alpha =0.95 , so, \alpha =0.05\) and \(z_{\alpha/2}=1.96\)

\(\sigma/ \sqrt n =6.7/\sqrt 36=1.117\)

So, \(95\%\) Confidence Interval (CI) for \(\mu\) is:

\(\bar x - z_{\alpha/2} *\frac{\sigma}{\sqrt n}<\mu<\bar x + z_{\alpha/2} *\frac{\sigma}{\sqrt n}\)

Or, \(23 - 1.96* 1.117<\mu<23 - 1.96* 1.117\)

Or,\(20.81<\mu<25.19\)

Practical Interpretation: Based on the sample data, a range of highly plausible values for mean sodium chloride concentration for seawater is \(20.81 cm^3<\mu<25.19cm^3\).

  1. The Energy Information Association claims that the mean monthly residential electricity consumption in your town is more than 874 kilowatt-hours (kWh). To test this claim, you find that a random sample of 64 residential customers has a mean monthly electricity consumption of 905 kWh. Assume the population standard deviation is 125 kWh. At α=0.05, do you have enough evidence to support the association’s claim?

Solution

Here, \(n=64, \bar x=905\ kWh ,\ \sigma= 125\ kWh\)

Hypothesis: \[H_0: \mu \le 874\] against, \[H_A: \mu>874\ (claim) \]

Test statistic \[z_{cal}=\frac{\bar x-\mu_0}{\sigma/\sqrt n}\] \[=\frac{905-874}{125/\sqrt 64}=1.98\] P-value approach

For right-tailed test, \(P\ value=P(Z>z_{cal})=P(Z>1.98)=1-\Phi(1.98)=0.024\)

Decision

Since, \(P \ value<\alpha\); so reject \(H_0\).

Practical Interpretation: Based on the sample data it can be concluded with 95% confidence that the association’s claim is true that is the mean monthly residential electricity consumption in your town is more than 874 kilowatt-hours (kWh).