please explain so i can understand. thank you!! In a recent year, 73% of 1st-yea
ID: 3055372 • Letter: P
Question
please explain so i can understand. thank you!!
In a recent year, 73% of 1st-year college students responding to a national survey identified “being very well-off financially” as an important personal goal. In a random sample of 200 of its 1st-year students, a state university finds that 132 students say that this goal is important.
State a 90% +4 confidence interval for the true proportion of 1st year students that say this goal is important:
We are [a]% confident that the [b] that identified “being very well-off financially” as an important personal goal is between [c]% and [d]%.
Test the hypothesis that the proportion of 1st-year students at this university who think this goal is important differs from the national value of 73% at the 0.10 significance level.
Ho: [e] Ha: [f]
Compute the test statistic:[g] and P-value:[h]
State your conclusion in the context of the problem:
We [i] the null hypothesis. There [j] (is/is not) statistical evidence that the proportion of 1st-year students at this university who think this goal is important differs from [k] at the 0.10 significance level.
Explanation / Answer
n = 200
p = 0.66
z-value of 94% CI = 1.8800
SE = sqrt(p*(1-p)/n)
= sqrt(0.66 * (1 -0.66)/200)
= 0.03350
ME = z*SE
= 1.88 * 0.03350
= 0.06297
Lower Limit = p - ME
= 0.66 - 0.06297
= 0.59703
Upper Limit = p + ME
= 0.66 + 0.06297
= 0.72297
94% CI (0.597 , 0.723 )
By using z = 1.64 at 90%
n = 200
p = 0.66
z-value of 95% CI = 1.6400
SE = sqrt(p*(1-p)/n)
= sqrt(0.66 * (1 - 0.66)/200)
= 0.03350
ME = z*SE
= 1.64 * 0.03350
= 0.05493
Lower Limit = p - ME
= 0.66 - 0.05493
= 0.60507
Upper Limit = p + ME
= 0.66 + 0.05493
= 0.71493
90% CI (0.6051 , 0.7149 )
We are 90% confident that the [b] that identified “being very well-off financially” as an important personal goal is between 60.51% and 71.49%
e)
H0 : P = 0.73
f)
Ha : P not equals to 0.73
g)
Test statistcs:
z = ( p - P)/sqrt(P * ( 1-p)/n)
= (0.66 - 0.73) / sqrt(0.73 * (1-0.73)/200)
= -2.2298
h)
P value = 0.0258
i)
We reject the null hypothesis. There is statistical evidence that the proportion of 1st-year students at this university who think this goal is important differs from the national value of 73% at the 0.10 significance level.
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