4. The LA times printed the headline "Most California Voters Support Legalizing Recreational Marijuana." That headline was based on a survey of 1879 Californians, with 1090 of them supporting Proposition 64 (which passed during the November 2016 election and opened up the state for legal cannabis sales starting in 2018). Use the survey results to construct a 95% confidence interval estimate for the percentage of Californians who support legalization of recreational marijuana. Is the claim in the headline supported by the poll findings?

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Answe question 4 only

5:46 PM Fri Mar 31
no
88
Part 3:
Sample size N= 2036
Sample proportion p^= 0.40
Confidence level is 0.95
9
Significance level a= 1-0.95
= 0,05
07+lab+110
★
Where, Za/2 is the critical Z
value at 0,05/2= 0.025
Use Excel function
"=ABS(Norm.S.INV(0.025))" to
obtain the Z critical value.
Formula: CI= p^ ± Za/2 p^(1-p^/n)
T
Yes, we can safely say that fewer than
50% of adults prefer to get their news
online.
Part 2. Estimating Population Parameters
3. In a Harris poll of 2036 adults, 40% said that they prefer to get their news online. Construct a 95%
confidence interval estimate of the percentage of adults who say that they prefer to get their news
online. Can we safely say that fewer than 50% of adults prefer to get their news online?
4. The LA times printed the headline "Most California Voters Support Legalizing Recreational Marijuana."
That headline was based on a survey of 1879 Californians, with 1090 of them supporting Proposition 64
(which passed during the November 2016 election and opened up the state for legal cannabis sales
starting in 2018). Use the survey results to construct a 95% confidence interval estimate for the
percentage of Californians who support legalization of recreational marijuana. Is the claim in the headline
supported by the poll findings?
What is a 95% confidence, interval,
estimate of the percentage of adults
who that they prefer to get their
say
news online is (0.3787,0.4213) or
(37.887%, 42.13%).
= 0.40_1.96 0.40(1-0,40)/2036
= 0.40 1.96 x 0,0108572
Chalkduster
= 0,40 -0.0212801
= (0.40-0,0212801, 0,40+0,0212801)
~ (0.3787, 0.4213)
ther
S
15 =
Part 4:
5G 89%
X
5. Using the 05 Freshman 15 data set found on Canvas, construct a 90% confidence interval estimate of the
mean weight change for the population. Write at least one sentence explaining the result.
Transcribed Image Text:5:46 PM Fri Mar 31 no 88 Part 3: Sample size N= 2036 Sample proportion p^= 0.40 Confidence level is 0.95 9 Significance level a= 1-0.95 = 0,05 07+lab+110 ★ Where, Za/2 is the critical Z value at 0,05/2= 0.025 Use Excel function "=ABS(Norm.S.INV(0.025))" to obtain the Z critical value. Formula: CI= p^ ± Za/2 p^(1-p^/n) T Yes, we can safely say that fewer than 50% of adults prefer to get their news online. Part 2. Estimating Population Parameters 3. In a Harris poll of 2036 adults, 40% said that they prefer to get their news online. Construct a 95% confidence interval estimate of the percentage of adults who say that they prefer to get their news online. Can we safely say that fewer than 50% of adults prefer to get their news online? 4. The LA times printed the headline "Most California Voters Support Legalizing Recreational Marijuana." That headline was based on a survey of 1879 Californians, with 1090 of them supporting Proposition 64 (which passed during the November 2016 election and opened up the state for legal cannabis sales starting in 2018). Use the survey results to construct a 95% confidence interval estimate for the percentage of Californians who support legalization of recreational marijuana. Is the claim in the headline supported by the poll findings? What is a 95% confidence, interval, estimate of the percentage of adults who that they prefer to get their say news online is (0.3787,0.4213) or (37.887%, 42.13%). = 0.40_1.96 0.40(1-0,40)/2036 = 0.40 1.96 x 0,0108572 Chalkduster = 0,40 -0.0212801 = (0.40-0,0212801, 0,40+0,0212801) ~ (0.3787, 0.4213) ther S 15 = Part 4: 5G 89% X 5. Using the 05 Freshman 15 data set found on Canvas, construct a 90% confidence interval estimate of the mean weight change for the population. Write at least one sentence explaining the result.
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