Question

Question Help Preliminary data analyses indicate that the assumptions for using pooled t-procedures are satisfied. Independent random samples of 102 plots with cropland and 100 plots with grassland Cropland Grassland x1 14.33x215.86 1 4.62 82-4.74 n1 -102 n2 100 ded the given summary statistics for the number of species of bird. At the 1% significance level do the data provide sufficient evidence to conclude that a difference exists in the me number of species in the two regions? Let μ1 be the mean number of species of bird in the cropland μ2 the mean number of species in the grassland, what are the correct hypotheses to test? Compute the test statistic. t(Round to three decimal places as needed.) Determine the critical value or values. Round to three decimal places as needed. Use a comma to separate answers as needed.) What is the conclusion of the hypothesis test? Since the test statistic ▼ in the rejection region, Ho Conclude that the data Vsufficient evidence that a difference exists in the mean number of species i n the two regions.

fill in the blank is 1. is not/is 2. do not reject/reject and 3. provide/do not provide

Answer 1

Solution :-

Given data :-

Mean, $\bar{x}_{1}= 14.33$

$\bar{x}_{2}= 15.86$

Standard deviation, $S_{1}= 4.62$

  $S_{2}= 4.74$

Sample size, $n_{1}= 102$

  $n_{2}= 100$

Level of significance, $\alpha =1$% =0.01

( i ) :-

Here, $\mu _{1}$ be the mean number of species of bird in the cropland and $\mu _{2}$ be the mean number of species of bird in the grassland.

The hypotheses to test is

Null hypothesis, $H_{0}:\mu _{1}=\mu _{2}$

Alternative hypothesis, $H_{0}:\mu _{1}\neq \mu _{2}$

Hence Option-D is correct answer.

( ii ) :-   

Now consider the test statistic, t

$t=\frac{(\bar{x_{1}}-\bar{x_{2}})} {\sqrt{\frac{S_{1}^2}{n_1}+\frac{S_{2}^2}{n_2}}}$

By substituting given values we get.

$t=\frac{(14.33-15.86)} {\sqrt{\frac{4.62^2}{102}+\frac{4.74^2}{100}}}$

$t=\frac{-1.53} {\sqrt{\frac{21.3444}{102}+\frac{22.4676}{100}}}$

$t=\frac{-1.53} {\sqrt{0.20925+0.224676}}$

$t=\frac{-1.53} {\sqrt{0.4339}}$

$t=\frac{-1.53} {0.6587}$

$t=-2.322$

Test statistic is, $t=-2.322$
( iii ) :-

Here we know that

The critical values at 1% confidence interval is  $\pm 2.33$

( iv ) :-

Here the t-value should be < -2.33 or >2.33.

Hence we can conclude that as , Since the test statistic is in the rejection region, Do not reject  $H_{0}$. Conclude that the data do not provide sufficient evidence that a difference exists in the mean number of species in the two regions.