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Trava [24]
3 years ago
12

Guys please please help me with this question

Mathematics
1 answer:
AleksandrR [38]3 years ago
4 0

Answer:

population in 1975 okkkkkkk

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In a simple random sample of 14001400 young​ people, 9090​% had earned a high school diploma. Complete parts a through d below.
ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has ​increased.

Step-by-step explanation:

Let <em>p</em> = proportion of young people who had earned a high school diploma.

A sample of <em>n</em> = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

\hat p=0.90

(a)

The standard error for the estimate of the percentage of all young people who earned a high school​ diploma is given by:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

Compute the standard error value as follows:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

       =\sqrt{\frac{0.90(1-0.90)}{1400}}\\

       =0.008

Thus, the standard error for the estimate of the percentage of all young people who earned a high school​ diploma is 0.0080.

(b)

The margin of error for (1 - <em>α</em>)% confidence interval for population proportion is:

MOE=z_{\alpha/2}\times SE_{\hat p}

Compute the critical value of <em>z</em> for 95% confidence level as follows:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the margin of error as follows:

MOE=z_{\alpha/2}\times SE_{\hat p}

          =1.96\times 0.0080\\=0.01568\\\approx1.6\%

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has​ increased, the hypothesis is defined as:

<em>H₀</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> = 0.80.

<em>Hₐ</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of <em>p</em>, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has ​increased.

8 0
3 years ago
A bottle of hand lotion is on sale for 2.25.if the price represents a 50percent discount from the original price,what is the ori
Flura [38]

Answer: $4.50


Step-by-step explanation: If it is 50% off then you take the price of $2.25 and double it. So 2 times $2.25 which is $4.50.

5 0
3 years ago
A rectangle has an area of 72 in². The length and the width of the rectangle are changed by a scale factor of 3.5.
vodka [1.7K]

Answer:

882 in^2

Step-by-step explanation:

4 0
3 years ago
How to simplify 2mg x 5m²k​
Damm [24]

hey, I'm not sure about this answer, but I think it's true, sorry

m(2g+5mk)

6 0
3 years ago
Given that A and B are true and X and Y are false, determine the truth value of the following proposition: ~[(A ⊃ Y) v ~(X ⊃ B)]
Anni [7]

Answer:

The value of the proposition is FALSE

Step-by-step explanation:

~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ ~X) v (B ⊃ X)]

Let's start with the smallest part: ~X. The symbol ~ is negation when X is true with the negation is false and vice-versa. In this case, ~X is true (T)

~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ T) v (B ⊃ X)]

Now the parts inside parenthesis:  (A ⊃ Y),(X ⊃ B),(A ≡ T) and (B ⊃ X). The symbol ⊃ is the conditional and A ⊃ Y is false when Y is false and A is true, in any other case is true. The symbol ≡ is the biconditional and A ≡ Y is true when both A and Y are true or when both are false.

(A ⊃ Y) is False (F)

(X ⊃ B) is True (T)

(A ≡ T) is True (T)

(B ⊃ X) is False (F)

~[(F) v ~(T)] ⋅ [~(T) v (F)]

The two negations inside the brackets must be taken into account:

~[(F) v F] ⋅ [F v (F)]

The symbol left inside the brackets v is the disjunction, and A v Y is false only  with both are false. F v (F) is False.

~[F] ⋅ [F]

Again considerating the negation:

T⋅ [F]

Finally, the symbol ⋅ is the conjunction, and A v Y is true only with both are true.

T⋅ [F] is False.

5 0
3 years ago
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