Math - Answers

Step 1. Name the test & check conditions Significance test for a single proportion Conditions We'll assume that the pennies were spun the same way (random) np [(40)(.5)=20>10%] [n(1-p)] [40(1-.5)=20>10] 3- 10n=400< total number of spins of pennies (unbounded) Step 2. State the hypothesis. P= true proportion of spins of pennies that come up heads. Ho: p=p0=0.5 Ha: P≠P0 Step 3 Calculate the test statistic and make a diagram. \[P_{0}=0.5\] \[x=10\] \[z=\frac{\hat{p}-\beta_{0}}{\sqrt{\frac{\rho_{0}|t-\rho_{0})}{n}}}\] \[n=40\] \[\hat{\overline{p}}=\frac{x}{n}=\frac{i0}{40}=0.25\] \[\overline{t}=\frac{.15-0.5}{\sqrt{\frac{(5)(.5)}{10}}}=-3.16227766\] \[\alpha=5\%\] then \(z^{*}=\pm1.96\) -196 0 1 1.96 \[-3.16277.6=z\] Step 4 Write the conclusions. Because (z\approx-3.162277<-1.96) we veject the null hypothesis Ho in favor of the alternate hypothesi's Ha If heads comes up 50% of the time when spinning pennies, then it would be very rove an to devations away from the mean (more more than 3 standard get 10 heads out of 40. Thus we have evidence to assert that heads do not cume up 50% of the time when spinning pennies. Step 3 Calculate the test statistic and make a diagram. Po = 0.5 x = 10 n = 40 p̂ = x/n = 10/40 = 0.25 z = (p̂ - Po) / sqrt(Po(1-Po)/n) = (0.25 - 0.5) / sqrt((0.5)(0.5)/40) = -3.16227766 α = 5% then z* = ±1.96 -1.96 0 1 1.96 -3.16277.6 = z Step 4 Write the conclusions. Because z ≈ -3.162277 < -1.96 we reject the null hypothesis Ho in favor of the alternate hypothesis Ha If heads comes up 50% of the time when spinning pennies, then it would be very rare (more than 3 standard deviations away from the mean) to get 10 heads out of 40. Thus we have evidence to assert that heads do not come up 50% of the time when spinning pennies. Identify the test (circle one): Significance test for one proportion (1-PropZTest) Significance test for the difference of two proportions (2-PropZTest) Significance test for a mean (T-Test) Significance test for the difference of two means (2-SampTTest) Data: Significance level Check the conditions: [x=\underline{5%}=.05] 1. The sample was obtained at random. Other: (depending on the test, circle and give the necessary values) 5 22/40--55 Pu 2. np0 = 40(1.5)=20>10√ \[n(1-p_{0})=40(-5)=10>10\cdot\] = = 40 \(10n=400 10 n(1-p₀) = (1027)(0.5) = 513.5 > 10 10n = 10270 < total number of adults Other: (depending on the test, circle and give the necessary values) p̂ = 544/1027 ≈ 0.527698 State the hypotheses: H₀: p = p₀ Hₐ: p > p₀ where (describe in words the appropriate symbol(s): p, p₀, p̂, μ, μ₀, μ̂) p = true proportion of adults who believe friends are important for success. Compute the test statistic, p-value, and label and complete the sketch: Test statistic: z = (p̂ - p₀) / sqrt(p₀(1-p₀)/n) ≈ 1.9035 P-value = 0.0285 Is the sample significant? Yes Conclusions: (Circle one) We reject/don't reject the null hypothesis. Reject the null hypothesis. Explain in context Because the P-value = 0.0285 < 0.05 = α, we reject the null hypothesis. If the proportion of adults who believe friends are important for success was equal to 50%, then it would be unlikely (probability 2.85%) to get a sample more extreme than 544 out of 1027. Thus we have enough evidence to assert that more than 50% of adults believe good friends are important for success.