Question

An oceanographer wants to test, on the basis of the mean of a random sample of size ? = 35 and at the 0.05 level of significance, whether the average depth ocean in a certain area is 72.4 fathoms, as has been recorded. What will she decide if she gets ?̅ = 73.2 fathoms, and she can assume from information gathered in similar studies that σ = 2.1 fathoms?

Answer 1

Solution:

Given:

Sample size = n = 35

Sample mean =

7=733

Level of significance = 0.05

Population Standard Deviation =

0 = 2.1

We have to test whether the average depth ocean in a certain area is 72.4 fathoms or not, since this statement is non-directional, we use two tailed test.

Thus we use following steps:

Step 1) State H₀ and H₁:

Η :μ = 72.4

Vs

H: # 72.4

Step 2) Find test statistic:

urlo

73.2 – 72.4 2.1/35

| 0.8 2.15.916080

10.8 0.354965

了=25

Step 3) Find z critical values:

Level of significance = 0.05

Since this is two tailed test, find Area = 0.05/2 = 0.025

Look in z  table for Area = 0.0250 or its closest area and find corresponding z value.

– -3.4 -3.3 -3.2 -3.1 -3.0 -2.9 -2.8 -2.7 -2.6 -2.5 -2.4 .00 .0003 .0005 .0007 .0010 .0013 .0019 .0026 .0035 .0047 .0062 .0082 .01 .0003 .0005 .0007 .0009 .0013 .0018 .0025 .0034 .0045 .0060 .0080 .0104 .0136 .0174 .0222 :0281 .0351 .02 .03 .04 .05 .06 .0003 .0003 .0003 .0003 .00b3 .0005 .0004 .0004 .0004 .00b4 .0006 .0006 .0006 .0006 .0006 .0009 .0009 .0008 .0008 .00 08 .0013 .0012 .0012 .0011 .0011 .0018 .0017 .0016 .0016 .00]5 .0024 .0023 .0023 .0022 .0021 .0033 .0032 .0031 .0030 .0099 .0044 .0043 ,0041 .0040 .0019 .0059 .0057 .0055 .0054 .0042 .0078 .0075 .0073 .0071 .009 .0102 .0099 .0096 .0094 .0091 .0132 .0129 .0125 .0122 .01I9 .0170 .0166 .0162 .0158 .0154 .0217 .0212 .0207 .0202 .017 .0274་ .02692620256.0250 .0344 .0336 .0329 .0322 .0314 .07 .08 .0003 .0003 .0004 .0004 .0005 .0005 .0008 .0007 .0011 .0010 .0015 .0014 .0021 .0020 .0028 .0027 .0038 .0037 .0051 .0049 .0068 .0066 .0089 .0087 .01160113 .0150 .0146 .0192 .0188 .0244 .0239 .0307 .0301 .09 .0002 .0003 .0005 .0007 .0010 .0014 .0019 .0026 .0036 .0048 .0064 .0084 .0110 .0143 .0183 .0233 .0294 ཀྐཱ -2.2 -2.1 -2.0 -1.9 -1.8 .0139 .0179 .0228 ཀྐ .0359

Area 0.0250 corresponds to -1.9 and 0.06

thus z critical value = -1.96

Since this is two tailed test, we have two z critical values: ( -1.96 , 1.96)

Step 4) Decision rule ( Rejection region)

Reject null hypothesis ,if z  test statistic value < z critical value = -1.96

or z  test statistic value> z critical value = 1.96   ,

otherwise we fail to reject H0.

Since  z  test statistic value = 2.25 > z critical value = 1.96 , we reject null hypothesis H0.

Step 5) Conclusion:

There is not sufficient evidence to conclude that: the average depth ocean in a certain area is 72.4 fathoms

that is : the average depth ocean in a certain area is different from 72.4 fathoms