Question

You plan to conduct a marketing experiment in which students are to taste one of two different brands of soft drink. Their task is to correctly identify the brand tasted. You select a random sample of 200 students and assume that the students have no ability to distinguish between the two brands. The probability is 90% that the sample percentage is contained within what symmetrical limits of the population percentage

Answer 1

Answer:

the probability is 90% that the sample percentage is contained within 45.5% and 54.5% symmetric limits of the population percentage.

Step-by-step explanation:

From the given information:

Sample size n = 200

The standard deviation for a sampling distribution for two brands are equally likely because the individual has no ability to discriminate between the two soft drinks.

∴

The population proportion $p_o$ = 1/2 = 0.5

NOW;

$\sigma _p = \sqrt{\dfrac{p_o(1-p_o)}{n}}$

$\sigma _p = \sqrt{\dfrac{0.5(1-0.5)}{200}}$

$\sigma _p = \sqrt{\dfrac{0.5(0.5)}{200}}$

$\sigma _p = \sqrt{\dfrac{0.25}{200}}$

$\sigma _p = \sqrt{0.00125}$

$\sigma _p = 0.035355$

However, in order to determine the symmetrical limits of the population percentage given that the z probability is 90%.

we use the Excel function as computed as follows in order to determine the z probability  = NORMSINV (0.9)

z value = 1.281552

Now the symmetrical limits of the population percentage can be determined as: ( 1.28, -1.28)

$1.28 = \dfrac{X - 0.5}{0.035355}$

1.28 × 0.035355 = X - 0.5

0.0452544= X - 0.5

0.0452544 + 0.5 = X

0.5452544 = X

X $\approx$ 0.545

X = 54.5%

$-1.28 = \dfrac{X - 0.5}{0.035355}$

- 1.28 × 0.035355 = X - 0.5

- 0.0452544= X - 0.5

- 0.0452544 + 0.5 = X

0.4547456 = X

X $\approx$ 0.455

X = 45.5%

Therefore , we can conclude that the probability is 90% that the sample percentage is contained within 45.5% and 54.5% symmetric limits of the population percentage.