Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t/2, (b) find the critical value z,/2, or (c) state that neither the normal distribution nor the t distribution applies.230, x 32.6 hg, s = 7.9 hg. The confidence level is 90%Here are summary statistics for randomly selected weights of newborn girls: nSelect the correct choice below and, if necessary, fill in the answer box to complete your choice.A.(Round to two decimal places as needed.)В.Za/2(Round to two decimal places as needed.)C. Neither the normal distribution nor the t distribution applies.

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Asked Nov 26, 2019
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Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t/2, (b) find the critical value z,/2, or (c) state that neither the normal distribution nor the t distribution applies.
230, x 32.6 hg, s = 7.9 hg. The confidence level is 90%
Here are summary statistics for randomly selected weights of newborn girls: n
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
(Round to two decimal places as needed.)
В.
Za/2
(Round to two decimal places as needed.)
C. Neither the normal distribution nor the t distribution applies.
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Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t/2, (b) find the critical value z,/2, or (c) state that neither the normal distribution nor the t distribution applies. 230, x 32.6 hg, s = 7.9 hg. The confidence level is 90% Here are summary statistics for randomly selected weights of newborn girls: n Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. (Round to two decimal places as needed.) В. Za/2 (Round to two decimal places as needed.) C. Neither the normal distribution nor the t distribution applies.

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Expert Answer

Step 1

The given information

Number of samples (n) =230

Mean of the samples =32.6

Standard deviation of the sample =7.9

Confidence interval =90%

α=1-0.9=0.1

Step 2

The correct option is b.

To find the critical value of z test is,

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0.1 0.05 from the table 1.64 0.05 =1.64 le

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Step 3
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=1.64 0.04 0.01 0.07 0.08 0 0.02 0.03 0.05 0.06 0.09 48405 46414 -0 50000 49601 49202 48803 48006 47608 47210 46812 -0.1 46017 45620 45224 44828 44433 44034 43640 43251 42858 42465 0.2 42074 41683 40905 40517 39743 39358 38974 38591 41294 40129 -0.3 36317 35942 38209 37828 37448 37070 36693 35569 35197 34827 -0.4 34458 32276 31561 34090 33724 33360 32997 32636 31918 31207 -0.5 30503 29460 .30854 30153 29806 29116 28774 28434 28096 27760 -0.6 26109 27093 27425 26763 26435 25785 25463 25143 24825 24510 -0.7 23885 24196 23576 23270 22965 22663 22363 22065 21770 21476 -0.8 21186 20897 20611 19766 20327 20045 19489 19215 18943 18673 0.9 17879 16354 16109 18406 18141 17619 17361 17106 16853 16602 -1 15386 14917 15866 15625 15151 14686 14457 14231 14007 13786 -1.1 .13136 12924 12100 13567 13350 12714 12507 12302 11900 11702 -1.2 10749 10383 10027 11507 11314 11123 10935 10565 10204 09853 -1.3 09680 09510 09342 09176 09012 08851 08534 08379 08226 08692 -1.4 07493 08076 07927 07780 07636 07353 07215 07078 .06944 06811 -1.5 06552 06057 05821 05705 06681 06426 06301 .06178 05938 05592 -1.6 -L.7 05050 04947 05155 05480 05370 05262 04846 04746 04648 04551 04363 03836 04457 04272 04182 04093 04006 03920 03754 03673 -1.8 03593 03515 03438 03362 .03288 03216 03144 03074 03005 02938 -1.9 02872 02442 02385 02807 .02743 .02680 02619 02559 02500 .02330

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