Question

Suppose heights of dogs, in inches, in a city are normally distributed and have a known population standard deviation of 7 inches and an unknown population mean. A random sample of 15 dogs is taken and gives a sample mean of 34 inches. Find the confidence interval for the population mean with a 99% confidence level.

Answer 1

Answer:

Step-by-step explanation:

We would use the t- distribution.

From the information given,

Mean, μ = 34 inches

Standard deviation, σ = 7 inches

number of sample, n = 15

Degree of freedom, (df) = 15 - 1 = 14

Alpha level,α = (1 - confidence level)/2

α = (1 - 0.99)/2 = 0.005

We will look at the t distribution table for values corresponding to (df) = 14 and α = 0.005

The corresponding z score is 2.977

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

34 ± 2.977 × 7/√15

= 34 ± 2.977 × 1.81

= 34 ± 5.39

The lower end of the confidence interval is 34 - 5.39 =28.61

The upper end of the confidence interval is 34 + 5.39 =39.39

Answer 2

Answer:

99% confidence interval for the population mean is [29.34 , 38.65].

Step-by-step explanation:

We are given that heights of dogs, in inches, in a city are normally distributed and have a known population standard deviation of 7 inches and an unknown population mean.

A random sample of 15 dogs is taken and gives a sample mean of 34 inches.

So, the pivotal quantity for 99% confidence interval for the population mean is given by;

            P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean = 34 inches

             $\sigma$ = population standard deviation = 7 inches

             n = sample of dogs = 15

             $\mu$ = population mean

So, 99% confidence interval for the average age, $\mu$ is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99

P(-2.5758 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.5758 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.99

P( $\bar X - 2.5758 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.5758 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X - 2.5758 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.5758 \times {\frac{\sigma}{\sqrt{n} }$  ]

                                                 = [ $34 - 2.5758 \times {\frac{7}{\sqrt{15} }$ , $34 + 2.5758 \times {\frac{7}{\sqrt{15} }$ ]

                                                 = [29.34 , 38.65]

Therefore, 99% confidence interval for the population mean is [29.34 , 38.65].