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Part 1 You have been asked to estimate the mean number of seals killed by a polar bear in one year. The Polar Bear tab of thi
Polar Bear Annual Seal Kills 四四四四四四 m 23 4 5678gp tea 219 0 347
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Polar Bear Annual Seal Kills 四四四四四四 m 23 4 5678gp tea 219 0 347
Part 1 You have been asked to estimate the mean number of seals killed by a polar bear in one year. The Polar Bear tab of thi
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Answer #1

Part 1:

Based on the given data,

1.

CORREL D M N O P Q Polar Bear 3 Mean - 2 3 Returns the average (arithmetic mean) of its arguments, which can be numbers or naCORREL Polar Bear X fx =AVERAGE(B2:B52 C D F G Annual Seal Kills 200 266 |=AVERAGE(B2:352 219 AVERAGE(number1, [number2], ...

The best estimate of population mean (\mu) would be the sample mean (\overline{x}). Hence, to estimate the mean no. of annual sea kills, we may make use of the given sample of size 51 and compute the  mean no. of annual sea kills in the sample.

We get \widehat{\mu }=\overline{x}=274.67

2. The 100 (1-\alpha) % CI for population mean \mu can be obtained by the formula:

\overline{x}\mp t_{\alpha ,n-1}\frac{s}{\sqrt{n}}

For \alpha =0.05,n = 51

letin TINV Returns the inverse of the Students t-distribution

-TINV(0.05,50 TINV(probability, deg_freedom)

We get the critical value of t as 2.009.

3. Substituting the values:

Mean 274.67 Standard deviation =stdev STDEV Estimates standard deviation based on a sample (ignores logical values and text i

Mean Standard deviation 274.67 STDEV(B2:B52 STDEV(number1, (number2], ...)

We get s = 53.71

274.67\mp 2.009\frac{53.71}{\sqrt{51}}

= (259.56, 289.78)

Part 2:

1. Let \mu denote the mean age at which the infant first speaks. We have to test:

H_{0}:\mu \leq 7 Vs H_{a}:\mu > 7

2. From the given data,

\overline{x}=7.83,s = 2.91,n=81

3. The appropriate statistical test to test the above hypothesis would be a one sample t test.

But before running this test, we must ensure that the data satisfies the assumptions of this test:

- The data is continuous - The observations are independent - The data is normally distributed - There are no outliers   

Assuming that all the assumptions are satisfied:

The test statistic is given by:

- tes

with critical region given by: t>t_{crit} for right tailed test.

a.

The critical value of t is given by:  

Using excel,

Etin TINV Returns the inverse of the Students t-distribution

ETINV(0.10,80 TINV(probability, deg_freedom)(................Since, excel function gives only the two tailed probabilities, we must convert the one tailed 0.05 into two tailed 0.05*2 = 0.10)

We get t_{0.10,80}=1.664

Substituting the values obtained in the test statistic:

t=\frac{7.83-7}{2.91/\sqrt{81}}

= 2.56

4. To compute the p-value:

=tdi TDIST Returns the Students t-distribution

=TDIST(2.56,80,1 TDIST(x, deg_freedom, tails)

We get p-value = 0.0062

5. Since, the p-value of the test 0.0062 < 0.05, we may reject H0. We may conclude that the data provides sufficient evidence to support the claim that the mean age at which the infant first speaks is greater than 7 months.

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