Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 403 green peas and 158 yellow peas. a. Construct a 95% confidence interval to estimate of the percentage of yellow peas. b. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yelow peas is not 25% do the results contradict expectations? a. Construct a 95% confidence interval Express the percentages in decimal form <p (Round to three decimal places as needed) b. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations? O Yes, the confidence interval does not include 0.25, so the true percentage could not equal 25% No, the confidence interval includes 0.25, so the true percentage could easily equal 25%

Answer 1

Let X = Event that the sample of offspring contained Yellow peas

By definition of proportion,

$\widehat{p}$ = No. of favorable outcomes / Total no. of outcomes

= x / n

= 158 / (158 + 403)

= 158 / 561

= 0.282

The 100(1-$\alpha$)% CI for true proportion can be constructed using the formula:

$\widehat{p}\mp Z_{1-\alpha/2}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

For $\alpha = 0.05$ , from Normal table, looking for the area 1- 0.05 /2 = 1 - 0.025 = 0.975:

.07 STANDARD NORMAL DISTRIBUTION: Table Values Represent AREA to the LEFT of the Z score. z .00 .01 .02 .03 .04 .05 .06 .08 .09 0.0 .50000 .50399 .50798 .51197 .51595 .51994 52392 .52790 .53188 .53586 0.1 .53983 .54380 .54776 .55172 .55567 55962 .56356 .56749 .57142 .57535 0.2 .57926 .58317 .58706 59095 .59483 59871 .60257 .60642 .61026 .61409 0.3 .61791 .62172 .62552 .62930 .63307 .63683 .64058 .64431 .64803 .65173 0.4 .65542 .65910 .66276 .66640 .67003 .67364 .67724 .68082 .68439 .68793 0.5 .69146 .69497 .69847 .70194 .70540 .70884 .71226 .71566 .71904 .72240 0.6 .72575 72907 .73237 .73565 .73891 .74215 .74537 .74857 .75175 .75490 0.7 .75804 .76115 .76424 .76730 .77035 .77337 .77637 .77935 .78230 .78524 0.8 .78814 .79103 79389 .79673 79955 .80234 .80511 .80785 .81057 .81327 0.9 .81594 .81859 .82121 .82381 .82639 .82894 .83147 .83398 .83646 .83891 1.0 .84134 .84375 .84614 .84849 .85083 .85314 .85543 .85769 .85993 .86214 1.1 .86433 .86650 .86864 .87076 .87286 .87493 .87698 .87900 .88100 .88298 1.2 .88493 .88686 .88877 .89065 .89251 .89435 .89617 .89796 .89973 90147 1.3 90320 .90490 90658 .90824 90988 91149 .91309 91466 .91621 .91774 1.4 91924 92073 .92220 92364 92507 .92647 92785 .92922 93056 93189 1.5 93319 .93448 .93574 .93699 93822 93943 .94062 .94179 94295 .94408 1.6 94520 94630 .94738 94845 94950 95053 95154 95254 .95352 95449 1.7 95543 .95637 .95728 .95818 .95907 .95994 .96080 .96164 .96246 .96327 1.8 96407 .96485 .96562 .96638 .96712 96784 .96856 .96926 .96995 97062 1.9 97128 97193 .97257 97320 97381 .97441 .97500 .97558 97615 97670 2.0 97725 .97778 .97831 .97882 .97932 .97982 98030 .98077 .98124 98169 2.1 .98214 .98257 .98300 98341 .98382 .98422 .98461 .98500 .98537 98574 2.2 98610 .98645 98679 98713 .98745 .98778 .98809 98840 98870 98899 2.3 98928 .98956 .98983 99010 99061 .99086 99111 99134 .99158 2.4 .99180 99202 99224 .99245 .99266 99286 .99305 .99324 .99343 .99361 99036

We get Z = 1.96.Substituting the values,

$\widehat{p}\mp Z_{1-\alpha/2}\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

$=0.282\mp (1.96)\sqrt{\frac{0.282(1-0.282)}{561}}$

= (0.244, 0.319)

The 95% CI for proportion of yellow peas is obtained as (0.244,0.319).