We are given that in a certain school district, it was observed that 32% of the students in the element schools were classified as only children (no siblings).

However, in the special program for talented and gifted children, 135 out of 347 students are only children.

Let p = population proportion of only children in the special program.

So, Null Hypothesis, $H_0$ : p = 32% {means that the proportion of only children in the special program is equal to the proportion for the school district}

Alternate Hypothesis, $H_A$ : p $\neq$ 32% {means that the proportion of only children in the special program is significantly different from the proportion for the school district}

The test statistics that would be used here One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of only children in the special program = $\frac{135}{347}$ = 0.39

n = sample of students = 347

So, test statistics = $\frac{0.39-0.32}{\sqrt{\frac{0.39(1-0.39)}{347} } }$

= 2.673

The value of z test statistics is 2.673.

Now, at 0.02 significance level the z table gives critical value of -2.3263 and 2.3263 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of only children in the special program is significantly different from the proportion for the school district.

4 0

3 years ago

How do I graph y=3 sin0+1

cluponka [151]

You graph it by the y axis and that have a number 1 and you graph it straight sideway

3 0

4 years ago

At the movie theater, child admission is $5.50 and adult admission is $8.60. On Saturday, 137 tickets were sold for a total sale

snow_tiger [21]

Answer:

The number of child tickets were sold is 75

Step-by-step explanation:

Given as :

The price of tickets for child admission = $5.50

The price of tickets for adult admission = $8.60

The total numbers of tickets sold = 137

The total price of tickets sold = $986.00

Let The number of tickets for child = C

The number of tickets for adults = A

So, According to question

A + C = 137

And 5.50 A + 8.60 C = 986

Solve the above equation

5.5 A + 5.5 C = 137 × 5.5

or, 5.5 A + 5.5 C = 753.5

Or, ( 5.5 A + 8.6 C) - ( 5.5 A + 5.5 C) = 986 - 753.5

Or, (5.5 A - 5.5 A) + (8.6 C - 5.5 C) = 232.5

Or, 0 + 3.1 C = 232.5

∴ C = $\frac{232.5}{3.1}$ = 75

And A = 137 - C = 137 - 75 = 62

Hence The number of child tickets were sold is 75. Answer

7 0

4 years ago