Gabrielle Burch 1460 Test Chapter 7 PostTest Y U SolutionWe

Gabrielle Burch 1460 Test: Chapter 7 Post-Test Y U

Solution

We know that the distribution of sample proportion is normal distribution if two conditions are satisfied.

np 10 and nq 10

Here, n = 100 and p = 23% = 0.23

q = 1 - p = 1 - 0.23 = 0.77

np = 100 * 0.23 = 23

nq = 100 * 0.77 = 77

Both are > 10 so the sampling distribution of sample proportion (p^) is approximately normal with mean is p and and standard deviation is sqrt(pq/n).

mean = 0.23

sd = sqrt((0.23*0.77)/100) = 0.0421

i) We have to find P(p^ < 0.23).

Convert p^ into z-score.

z = (p^ - p) / sd

z = (0.23 - 0.23) / 0.0421 = 0

That is now we have to find P(Z < 0).

By using EXCEL syntax is,

=NORMSDIST(z)

where, z is test statistic.

P(Z < 0) = 0.5

b) We have to find P(0.24 < p^ < 0.3).

Convert 0.24 and 0.3 into z-score.

z = (0.24 - 0.23) / 0.0421 = 0.2376

z = (0.3 - 0.23) / 0.0421 = 1.6634

That is now we have to find P(0.2376 < Z < 1.6634) = P(Z < 1.6634) - P(Z < 0.2376)

By using EXCEL syntax is,

=NORMSDIST(z)

where, z is test statistic.

P(0.2376 < Z < 1.6634) = 0.9519 - 0.5939 = 0.36

iii) We have to find P(p^ > 0.24).

Convert p^ into z-score.

z = (p^ - p) / sd

z = (0.24 - 0.23) / 0.0421 = 0.2376

That is now we have to find P(Z > 0.2376).

By using EXCEL syntax is,

=1 - NORMSDIST(z)

where, z is test statistic.

P(Z > 0.2376) = 0.41

iv) n = 400

p^ = 0.23

sd = 0.0210

We have to find P(p^ < 0.28).

Convert p^ into z-score.

z = (p^ - p) / sd

z = (0.28 - 0.23) / 0.0210 = 2.3762

That is now we have to find P(Z < 2.3762).

By using EXCEL syntax is,

=NORMSDIST(z)

where, z is test statistic.

P(Z < 2.3762) = 0.99

v) We have to find P(0.24 < p^ < 0.3).

Convert 0.24 and 0.3 into z-score.

z = (0.24 - 0.23) / 0.0210 = 0.4752

z = (0.3 - 0.23) / 0.0421 = 3.3267

That is now we have to find P(0.4752 < Z < 3.3267) = P(Z < 3.3267) - P(Z < 0.4752)

By using EXCEL syntax is,

=NORMSDIST(z)

where, z is test statistic.

P(0.2376 < Z < 1.6634) = 0.9996 - 0.6827 = 0.32