Some people claim that they can tell the difference between a diet soda and a regular soda in the first sip. a researcher wanting to test this claim randomly sampled 80 such people. he then filled 80 plain white cups with soda, half diet and half regular through random assignment, and asked each person to take one sip from their cup and identify the soda as diet or regular. 53 participants correctly identified the soda. (a) do these data provide strong evidence that these people are able to detect the difference between diet and regular soda, in other words, are the results significantly better than just random guessing? (b) interpret the p-value in this context.

The difference between a diet soda and a regular soda illustrates normal distribution.

The given parameters are:

--- the sample size

--- the population

--- the proportion to put on half diet

So, the null and the alternate hypotheses are:

Calculating the proportion of individuals in the sample on half diet

Calculate standard error

So, we have:

The z-score is then calculated as:

So, we have:

Calculate the p-value

From z-score of probabilities, we have:

The p-value is less than .

So, we reject the null hypothesis that and accept the alternate hypothesis.

(b) Interpret the p-value

In (a), we have:

So, the interpretation is:

The probability of observing 53 or more claims out of 80 claims is 0.0018

Read more about normal probabilities at:

link

Part a)

p = population proportion of correct guesses

Null Hypothesis: p = 0.5

Alternative Hypothesis: p > 0.5

The claim that people can tell the difference of the two drinks is in the alternative hypothesis p > 0.5 meaning that it's more than just luck at play here. Saying p = 0.5 is basically saying there's a coin toss to determine the guess.

x = 53, n = 80

phat = x/n = 53/80 = 0.6625

SE = sqrt(p*(1-p)/n) = sqrt(0.5*(1-0.5)/83) = 0.05488212999484

z = (phat-p)/(SE)

z = (0.6625-0.5)/(0.05488212999484)

z = 2.96089091322218

z = 2.96

Use a table or calculator to find that

P(Z < 2.96) = 0.9985

So,

P(Z > 2.96) = 1-P(Z < 2.96)

P(Z > 2.96) = 1-0.9985

P(Z > 2.96) = 0.0015

The p-value is approximately 0.0015. This value is less than many alpha values commonly used (such as 0.01 or 0.05) so we reject the null hyptohesis that p = 0.5 and accept that p > 0.5 is true. So the participants aren't randomly guessing. The results are significantly better than random guesses.

Part b)

The p value 0.0015 is the probability of observing 53 or more claims out of 80 claims, under the assumption that the null hypothesis is true. This p value is very very small, so there is a small chance that the null hypothesis is correct. The smaller the p value, the more likely we reject the null. The general rule is that if the p value is smaller than the significance level alpha, we reject the null. Your textbook will state what alpha is equal to. If not, then the default is alpha = 0.05

idk this answer sorry cant on this one.

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.