Answer the question please

A survey of 700 adults from a certain region asked, "What do you buy from your mobile device?" The results indicated that 59%

Kryger [21]

Answer:

a) the test statistic z = 1.891

the null hypothesis accepted at 95% level of significance

b) the critical values of 95% level of significance is zα =1.96

c) 95% of confidence intervals are (0.523 ,0.596)

Step-by-step explanation:

A survey of 700 adults from a certain region

Given sample sizes $n_{1} = 400 and n_{2} = 300$

Proportion of mean $p_{1} = \frac{236}{400} = 0.59 and p_{2} = \frac{156}{300} = 0.52$

Null hypothesis H0 : assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

Alternative hypothesis H1:- p1 ≠ p2

a) The test statistic is

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }$

where $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

on calculation we get p = 0.56

now q =1-p = 1-0.56=0.44

$Z = \frac{p_{1} -p_{2} }{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} )} } }\\ =\frac{0.56-0.52}{\sqrt{0.56X0.44}(\frac{1}{400}+\frac{1}{300} }$

after calculation we get z = 1.891

b) The critical value at 95% confidence interval zα = 1.96 (from z-table)

The calculated z- value < the tabulated value

therefore the null hypothesis accepted

conclusion:-

assume that there is no significant difference between males and women reported they buy clothing from their mobile device

p1 = p2

c) 95% confidence intervals

The confidence intervals are P± 1.96(√PQ/n)

we know that = $p = \frac{n_{1}p_{1} +n_{2}p_{2} }{n_{1}+n_{2}}= \frac{400X0.59+300X0.52}{700}$

after calculation we get P = 0.56 and Q =1-P =0.44

Confidence intervals are ( P- 1.96(√PQ/n), P+ 1.96(√PQ/n))

now substitute values , we get

( 0.56- 1.96(√0.56X0.44/700), 0.56+ 1.96(0.56X0.44/700))

on simplification we get (0.523 ,0.596)

Therefore the population proportion (0.56) lies in between the 95% of confidence intervals (0.523 ,0.596)

3 0

3 years ago

NEED HELP ASAP!!!!!!!!

Ymorist [56]

Segment EG is half the length of segment BH because of the Midsegment theorem

5 0

3 years ago

Melissa has a collection of different colored hair bows. Some of the hair bows are big, and some of the hair bows are small. The

Brilliant_brown [7]

Answer:

7/55 or 12.727272727273%

Step-by-step explanation:

7/55 or 12.727272727273%

7 big and pink

so 7/?

?= amount of bows

and them up and you get 55

7/55

4 0

2 years ago

Why does a radical function with an even index only appear on one sode of the x-axis while a radical woth an odd index appears o

Vladimir79 [104]

Consider the equation y = x^2. No matter what x happens to be, the result y will never be negative even if x is negative. Example: x = -3 leads to y = x^2 = (-3)^2 = 9 which is positive.

Since y is never negative, this means the inverse x = sqrt(y) has the right hand side never be negative. The entire curve of sqrt(x) is above the x axis except for the x intercept of course. Put another way, we cannot plug in a negative input into the square root function for this reason. This similar idea applies to any even index such as fourth roots or sixth roots.

Meanwhile, odd roots such as a cube root has its range extend from negative infinity to positive infinity. Why? Because y = x^3 can have a negative output. Going back to x = -3 we get y = x^3 = (-3)^3 = -27. So we can plug a negative value into the cube root to get some negative output. We can get any output we want, negative or positive. So the range of any radical with an odd index is effectively the set of all real numbers. Visually this produces graphs that have parts on both sides of the x axis.

3 0

3 years ago

A production process is checked periodically by a quality control inspector. the inspector selects simple random samples of 30 f

777dan777 [17]

Answer:

Population Mean = 2.0

Population Standard deviation = 0.03

Step-by-step explanation:

We are given that the inspector selects simple random samples of 30 finished products and computes the sample mean product weight.

Also, test results over a long period of time show that 5% of the values are over 2.1 pounds and 5% are under 1.9 pounds.

Now, mean of the population is given the average of two extreme boundaries because mean lies exactly in the middle of the distribution.

So, Mean, $\mu$ = $\frac{1.9+2.1}{2}$ = 2.0

Therefore, mean for the population of products produced with this process is 2.

Since, we are given that 5% of the values are under 1.9 pounds so we will calculate the z score value corresponding to a probability of 5% i.e.

z = -1.6449 {from z % table}

We know that z formula is given by ;

$Z = \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

-1.6449 = $\frac{1.9 - 2.0}{\frac{\sigma}{\sqrt{n} } }$ ⇒ $\frac{\sigma}{\sqrt{n} } = \frac{-0.1}{-1.6449}$

⇒ $\sigma =$ 0.0608 * $\sqrt{30}$ {as sample size is given 30}

⇒ $\sigma$ = 0.03 .

Therefore, Standard deviation for the population of products produced with this process is 0.0333.

7 0

3 years ago

Answer the question please ​

Answer the question please