Grey-Seal Nursing. The average lactation (nursing) period of all earless seals is 23 days. Grey seals are one of several types of earless seals. The length of time that a female grey seal nurses her pup are studied by S. Twiss et al. in the article "Variation in Female Grey Seal (Halichoerus grypus) Reproductive Performance Correlates to Proactive-Reactive Behavioural Types" (PLOS ONE 7(11): e49598. doi:10.1371/journal. pone.0049598). A sample of 14 female grey seals had the following lactation periods, in days.

20.2	20.9	20.6	23.6	19.6	15.9	19.8
15.4	21.4	19.5	17.4	21.9	22.3	16.4

At the significance level, do the data provide sufficient evidence to conclude that the mean lactation period of grey seals differs from 23 days? Assume that the population standard deviation is days. (Note: The sum of the data is days.)

given,

The conditions for normal distribution:-

• The shape of the distribution must be bell-shaped (symmetric).
• Probability plot: All points must be closer to a straight line

Now, Construct the normal probability plot by using MINITAB.
MINITAB procedure:
Step 1: Choose Graph $>$ Probability Plot.
Step 2: Choose Singles, and then click OK.
Step 3: In Graph variables, enter the column of Period.
Step 4: Click OK.

Observation:

According to the probability plot of the period, the observations are closer to a straight line.

Therefore, according to the given graph, the distribution of the period is approximately normally distributed.

Step 1:
State the null and alternative hypotheses:
The null hypothesis is:

$H_0:\mu =23\text{days}$

It is, that the mean lactation period of grey seals does not differ from 23 days.
An alternative hypothesis is :

$H_a:\mu \ne 23\text{days}$

It is, the mean lactation period of grey seals differs from 23 days.

Step 2: Decide on a significance level
Here, the significance level is, $\alpha =0.05$.

Step 3: Now, Compute the value of the test statistic by using the MINITAB.

MINITAB procedure:
Step 1: Choose State $>$Basic Statistics$>$ 1-Sample Z.
Step 2: In Samples, in Column, enter the column of Period.
Step 3: In Standard deviation, enter$3$.
Step 4: Perform hypothesis test, enter the test mear as
Step 5: Check Options, enter the confidence level as
Step 6: Choose not equal in alternative.
Step 7: Click OK in all dialogue boxes.

Step 4:

The critical value is:

From "Table ll Areas under the standard normal curve" the required value of $Z_{\frac{a}{2}}$with $95\%$ confidence level is $\pm 1.96$.

Now, Find P.value:

From the MINITAB output, the P.value is $0$.

Step 5:

The critical value approach:

The rejection rule is:

If the test statistic value falls in the rejection region, reject the null hypothesis $H_0$.

If the test statistic value does not fall in the rejection region, then does not reject the null hypothesis $H_0$

Here, the test statistic value falls in the rejection region.

Thus, the null hypothesis is rejected at the $5\%$level.

Hence, it is concluded that the test results are statistically significant at a $5\%$level of significance.

If$P\le \alpha$,

and reject the null hypothesis.

Here, the P-value is $0$, which is less than the significance level.

That is,

$P(=0)\le \alpha (=0.05)$.

Thus, the null hypothesis is rejected at the $5\%$ level.

Hence, it is concluded that the test results are statistically significant at a $5\%$ level of significance.

The data provided is sufficient evidence to conclude that the mean location period of grey seals differs from $23daysata5\%$level.