A person recently read that 84% of cat owners are women. How large a sample should the researcher take if she wishes to be 90% c
1 answer:
Answer:
405
Step-by-step explanation:
To find sample size, use the following equation, where n = sample size, za/2 = the critical value, p = probability of success, q = probability of failure, and E = margin of error.
The values that are given are p = 0.84 and E = 0.03.
You can solve for the critical value which is equal to the z-score of (1 - confidence level)/2. Use the calculator function of invNorm to find the z-score. The value will given with a negative sign, but you can ignore that.
(1 - 0.9) = 0.1/2 = 0.05
invNorm(0.05, 0, 1) = 1.645
You can also solve for q which is 1 - p. For this problem q = 1 - 0.84 = 0.16
Plug the values into the equation and solve for n.
Round up to the next number, giving you 405.
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Answer:
(a) The correct answer is P (CBM) = 0.79.
(b) The probability of selecting an American female who is not red-green color-blind is 0.996.
(c) The probability that neither are red-green color-blind is 0.9263.
(d) The probability that at least one of them is red-green color-blind is 0.0737.
Step-by-step explanation:
The variables CBM and CBW are denoted as the events that an American man or an American woman is colorblind, respectively.
It is provided that 79% of men and 0.4% of women are colorblind, i.e.
P (CBM) = 0.79
P (CBW) = 0.004
(a)
The probability of selecting an American male who is red-green color-blind is, 0.79.
Thus, the correct answer is P (CBM) = 0.79.
(b)
The probability of the complement of an event is the probability of that event not happening.
Then,
P(not CBW) = 1 - P(CBW)
= 1 - 0.004
= 0.996.
Thus, the probability of selecting an American female who is not red-green color-blind is 0.996.
(c)
The probability the woman is not colorblind is 0.996.
The probability that the man is not color- blind is,
P(not CBM) = 1 - P(CBM)
= 1 - 0.004
= 0.93.
The man and woman are selected independently.
Compute the probability that neither are red-green color-blind as follows:
Thus, the probability that neither are red-green color-blind is 0.9263.
(d)
It is provided that a one man and one woman are selected at random.
The event that “At least one is colorblind” is the complement of part (d) that “Neither is Colorblind.”
Compute the probability that at least one of them is red-green color-blind as follows:
Thus, the probability that at least one of them is red-green color-blind is 0.0737.
Answer:
a. two-tailed test
b. reject H0 if Z>1.96 or Z<-1.96
c. Z = 1.73
d. we have failed to reject the null hypothesis
e. P-value = 0.0418
Step-by-step explanation:
We will use a Z test to resolve this, an it will be a two-tailed test because the hypothesis statements are not indicating a specific direction for the significant difference (H0 : μ1 = μ2 ; H1 : μ1 ≠ μ2), this also means that the significanclevel will be divided between the both tails (2.5% en each tail for the rejection regions). See attached drawing for reference.
We need to find our critical value:
Zα/2 = Z(0.05/2) = 0.025
If we look for 0.025 in a Z table we will find that the critical value is 1.96 to the right, and by symmetry -1.96 to the left. So our decision rule will be to reject H0 if Z>1.96 or Z<-1.96
The Z test will be done using the next equation:
Z = (x⁻1 - x⁻2) - (μ1 - μ2) / √( σ²1/n1 + σ²2/n2
Because we are testing the null hypothesis we know that μ1 - μ2 must be zero if they are supposed to be equal (H0 : μ1 = μ2), so we calculate as follows:
Z = (100.5 - 98.8) - (0) / √( (3.4)²/38 + (5.8)²/51 = 1.7/0.9817 = 1.73
Z<1.96, therefore we have failed to reject the null hypothesis
The P-value for this test would be represented by the probability of Z being greater than 1.73, so we can look for it in any Z table, finding that its value is 0.0418, and because P-value>0.025 we again confirm that we don't have evidence statistically significant to reject the null hypothesis
