Given:

The average lifespan of a golden retriever is a set to be 9.6 years in the random sample of 15 golden retriever the following life span were found.

From tbe given data

Sample size, \(\displaystyle{n}={15}\)

\(\displaystyle\sum{x}={145.6}\)

Sample mean, \(\displaystyle\overline{{{x}}}=\sum\frac{{x}}{{n}}=\frac{{145.6}}{{15}}={9.7067}\)

Sample standard deviation, \(\displaystyle{s}={1.7689}\)

Population mean, \(\displaystyle\mu={9.6}\)

Hypothesis test:

The null and alternative hypothesis is

\(\displaystyle{H}_{{0}}:\mu={9.6}\)

\(\displaystyle{H}_{{a}}:\mu\ne{9.6}\)

Test statistics is

\(\displaystyle{z}={\frac{{\overline{{{x}}}-\mu}}{{\frac{\sigma}{\sqrt{{n}}}}}}={\frac{{{9.7067}-{9.6}}}{{\frac{{1.7689}}{\sqrt{{{15}}}}}}}={0.2336}\)

\(\displaystyle\therefore{z}={0.23}\)

p-value for two tailed:

p-value \(\displaystyle={2}{p}{\left({z}{>}{0.23}\right)}\)

\(\displaystyle={2}\times{0.409046}\)...(from z-table)

\(\displaystyle={0.8181}\stackrel{\sim}{=}{0.8187}\)

\(\displaystyle\therefore\) p-value \(\displaystyle={0.8187}\)

The average lifespan of a golden retriever is a set to be 9.6 years in the random sample of 15 golden retriever the following life span were found.

From tbe given data

Sample size, \(\displaystyle{n}={15}\)

\(\displaystyle\sum{x}={145.6}\)

Sample mean, \(\displaystyle\overline{{{x}}}=\sum\frac{{x}}{{n}}=\frac{{145.6}}{{15}}={9.7067}\)

Sample standard deviation, \(\displaystyle{s}={1.7689}\)

Population mean, \(\displaystyle\mu={9.6}\)

Hypothesis test:

The null and alternative hypothesis is

\(\displaystyle{H}_{{0}}:\mu={9.6}\)

\(\displaystyle{H}_{{a}}:\mu\ne{9.6}\)

Test statistics is

\(\displaystyle{z}={\frac{{\overline{{{x}}}-\mu}}{{\frac{\sigma}{\sqrt{{n}}}}}}={\frac{{{9.7067}-{9.6}}}{{\frac{{1.7689}}{\sqrt{{{15}}}}}}}={0.2336}\)

\(\displaystyle\therefore{z}={0.23}\)

p-value for two tailed:

p-value \(\displaystyle={2}{p}{\left({z}{>}{0.23}\right)}\)

\(\displaystyle={2}\times{0.409046}\)...(from z-table)

\(\displaystyle={0.8181}\stackrel{\sim}{=}{0.8187}\)

\(\displaystyle\therefore\) p-value \(\displaystyle={0.8187}\)