Question

Question 6 Find the critical value(s) that would be used to test eachdaim. If there are two critical values in your answer, separate them by a comma with the smaller one listed first. Charts • CVs for Normal Distribution .t-chart • Chi Square Chart Claim: > 15 Other Information: unknown, n = 10 Level of significance: a 0.05 Critical Value(s): Claim: p=0.54 level of significance: a = 0.10 Critical Value(s): Claim: o = 200 Other information: n=21 Level of significance a-0.01 Critical Value(s) Goodness of fit test with 7 categories. Level of significance: -0.05 Critical Valuels): app.honorlock.com is sharing your scree

Answer 1

Solution:

We have to find the critical values for following each of the claim:

Part 1)

Given: Claim:$\mu > 15$ , $\sigma$ is unknown , Sample size = n = 10 , Level of significance = $\alpha = 0.05$

Since $\sigma$ is unknown and sample size n =10 is small , we use t distribution to find t critical value.

Here claim is right tailed, so we use one tailed test.

Thus look in t table for df = n -1 = 10 - 1 = 9 and one tail area = 0.05 and find t critical value.

t Table cum. probt 50 one-tail) 0.50 two-tails 1.00 .76 0.25 0.50 t 80t 86 0.20 0.15 0.40 0.30 to 0.10 0.20 0.05 0.10 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 -0.000 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.7060 0.70 1.376 1.061 0.978 0.941 0.920 0.906 0.896 .889 0.882 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 6.14 2.920 2.153 2. 32 2.015 1.643 1.95 1.860 1.833

Thus t critical value = 1.833

Part 2)

Given: Claim : p = 0.54 Level of significance = $\alpha = 0.10$

Since claim is non-directional and this is test for proportion, thus we use two tailed z test.

Thus find Area = $\alpha/2 = 0.10/2=0.05$

look in z table for Area = 0.0500 or its closest area and find z value

1 -3.4 -3.3 -3.2 -3.1 -3.0 -2.9 -2.8 -2.7 -2.6 -2.5 -2.4 -2.3 -2.2 -2.1 -2.0 -1.9 -1.8 -1.7 -1.6 -1.5 ,00 10003 .0005 .0007 .0010 .0013 .0019 .0026 .0035 .0047 .0062 .0082 .0107 .0139 .0179 .0228 .0287 .0359 .0446 0548 10668 .01 .02 ,0003 ,0003 .0005 .0005 .0007 ,0006 ,0009 ,0009 .0013 .0013 .0018 .0018 .0025 .0024 .0034 .0033 .0045 .004 .0060 .0059 20080 .0078 .0104 .0102 .0136 .0132 .0174 .0170 .0222 .0217 .0281 .0274 .0351 .0344 .0436 .0427 .05370526 .0655 .0643 .03 .04 .06 ,0003 .00b3 .0b03 ,0003 ,0004 1.0004 .0004 .0004 ,0006 .0006 .0006 .0006 ,0009 ,0008 ,0008 ,0008 .0012 .0012 .0011 .0011 .0017 .0016 .0016 .0015 ,0023 .0023 .0022 .0021 .0032 .0031 .0030 .0029 .0043 .0041 0040 .0039 0057 1.0054 .0052 .0075 | .0073 .001.0069 .0099 20096 .0094 .0091 .0129 .0125 2012 .0119 .0166 .0162 ,018 .0154 .0212 .0407 .0202 .0197 .0268 .0262 .0256 .0250 .0336 .029 .0322 1.0314 ,0418 .0409 10401 .0392 .0516 .0505 .0495 .0485 ,0630 .0618 .0606 .0594 .07 .0003 .0004 .0005 .0008 .0011 .0015 .0021 .0028 .0038 .0051 .0068 .0089 .0116 .0150 .0192 .0244 .0307 .0384 .0475 ,0582

Area 0.0500 is in between 0.0495 and 0.0505 and both the area are at same distance from 0.0500

Thus we look for both area and find both z values

Thus Area 0.0495 corresponds to -1.65 and 0.0505 corresponds to -1.64

Thus average of both z values is : ( -1.64+ - 1.65) / 2 = -1.645

Thus Z = -1.645

Since this is two tailed test, we use two z critical values.

Thus z critical values are: ( -1.645 , 1.645 )

Part 3)

Given: Claim: $\sigma ^{2}=200$ , Sample size = n = 21 , Level of significance = $\alpha = 0.01$

Since claim is non-directional and this is test for Variance, we use two tailed Chi-square test of variance.

Thus find Area = $\alpha /2= 0.01/2=0.005$ and $1-\alpha /2= 1-0.005=0.995$

df = n - 1 = 21-1 =20

Look in Chi-square table for df =20 and Area = 0.995 and Area = 0.005 and find corresponding critical values.

Thus we get:

2995 X2975 X250 X2100 x2050 X2010 x2005 7979 0.000 0.010 3 2 2s | 33 10.597 12.38 14.460 16.750 18.518 20.78 X2990 0.000 0.020 0.115 0.297 0.554 0.872 1.239 1.646 2.088 2.558 3.053 3.571 4.107 4.660 5.229 5.812 6.408 7.015 7.633 8.200 0.001 0.051 0.216 0.484 0.831 1.237 1.690 2.180 2.700 3.247 3.816 4.404 5.009 5.629 6.262 6.908 7.564 8.231 8.907 .591 0.004 0.103 0.352 0.711 1.145 1.635 2.167 2.733 3.325 3,940 4.575 5.226 5.892 6.571 7.261 7.962 8.672 9.390 10.117 10.801 X2300 0.016 0.211 0.584 1.064 1.610 2.204 2.833 3.490 4.168 4.865 5.578 6.304 7.042 7.790 8.547 9.312 10.085 10.865 11.651 12.443 2.706 4.605 6.251 7.779 9.236 10.645 12.017 13.362 14.684 15.987 17.275 18.549 19.812 21.064 22.307 23.542 24.769 25.989 27.204 28.412 X2025 3.841 5.024 5.991 7.378 7.815 9.348 9.488 11.143 11.070 12.833 12.592 14.449 14.067 16.013 15.507 17.535 16.919 19.023 18,307 20.483 19.675 21.920 21.026 23.337 22.362 24.736 23.685 26.119 24.996 27.488 26.296 28.845 27.587 30.191 28.869 31.526 30.144 32.852 31.410134.TTU 6.635 9.210 11.345 13.277 15.086 16.812 18.475 20.090 21.666 23.209 24.725 26.217 27.688 29.141 30.578 32.000 33.409 34.805 36.191 37.360 21.55 23.589 25/188 26.757 2.603 3.04 3.555 4.055 4.cb1 5.112 5.67 6.265 35 718 37 156 381582 39.997 814 207.434 9

Chi-square left tail critical value = 7.434

Chi-square right tail critical value = 39.997

Thus critical values are: ( 7.434 , 39.997 )

Part 4)

Given: Goodness of fit test with 7 categories and Level of significance = $\alpha = 0.05$

Thus df = k - 1 = 7 - 1 = 6

Look in Chi-square table for df = 6 and right tail area = 0.05 and find critical value.

xii95 x900 0.000 0.010 0.072 0.207 0.412 x95 x950 0.000 0.001 0.004 0.020 0.051 0.103 0.115 0.216 0.352 0.297 0.484 0.711 0.554 0.831 | 1.145 21.237 16 - . .- 0.016 0.211 0.584 1.064 1.610 2201 xioo 2.706 4.605 6.251 7.779 9.236 - UPISI Xioso जला 5p91 7.815 9.188 11.070 12.592 ama

Chi-square critical value = 12.592

Thus critical value = 12.592