Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

In a recent survey, 10 percent of the participants rated Pepsi as being \"concer

ID: 3179520 • Letter: I

Question

In a recent survey, 10 percent of the participants rated Pepsi as being "concerned with my health." PepsiCo's response included a new "Smart Spot" symbol on its products that meet certain nutrition criteria, to help consumers who seek more healthful eating options.

Suppose a follow-up survey shows that 51 of 400 persons now rate Pepsi as being "concerned with my health."

At = .05, would a follow-up survey showing that 51 of 400 persons now rate Pepsi as being “concerned with my health” provide sufficient evidence that the percentage has increased?

In a recent survey, 10 percent of the participants rated Pepsi as being "concerned with my health." PepsiCo's response included a new "Smart Spot" symbol on its products that meet certain nutrition criteria, to help consumers who seek more healthful eating options.

Explanation / Answer

Solution:-

p = 51/400

p = 0.1275

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.10

Alternative hypothesis: P > 0.10

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.015

z = (p - P) /

z = 1.833

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is more than 1.833. We use the Normal Distribution Calculator to find P(z > 1.833) = 0.0334

Thus, the P-value = 0.0334

Interpret results. Since the P-value (0.0334) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we have sufficient evidence that the percentage has increased.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote