Question 1183827
Here's the solution:

**a. Null and Alternative Hypotheses:**

* **Null Hypothesis (H0):** The yearly consumption of soft drinks per person is 52 gallons. (μ = 52)
* **Alternative Hypothesis (H1):** The yearly consumption of soft drinks per person is *not* 52 gallons. (μ ≠ 52)  This is a two-tailed test.

**b. Difference between Degrees of Freedom and Sample Size:**

* Degrees of freedom (df) = Sample size (n) - 1 = 50 - 1 = 49
* Absolute difference = |n - df| = |50 - 49| = 1

**c. Product of Proportions:**

* People who consumed soft drinks = 50 - 8 = 42
* Proportion who consumed soft drinks = 42/50 = 0.84
* Proportion who did not consume soft drinks = 8/50 = 0.16
* Product of proportions = 0.84 * 0.16 = 0.1344

**d. Total Yearly Consumption:**

* Total consumption = Sample mean * Sample size
* Total consumption = 56.3 gallons/person * 50 people = 2815 gallons

**e. Test Statistic:**

Since the population standard deviation is known, we use a z-test.

z = (sample mean - population mean) / (population standard deviation / √sample size)
z = (56.3 - 52) / (3.5 / √50)
z = 4.3 / (3.5 / 7.071)
z ≈ 4.3 / 0.495
z ≈ 8.69

**f. P-value:**

Because this is a two-tailed test, we need to find the area in *both* tails of the standard normal distribution that is beyond our calculated z-score.

* Since the z-score is very large (8.69), the p-value will be extremely small, essentially close to zero.  Most z-tables won't go out this far; a calculator or statistical software is best.

**g. Validity of the Researcher's Claim:**

* Significance level (α) = 0.05
* Our p-value is essentially 0, which is *much* less than 0.05.

Since the p-value is less than the significance level, we reject the null hypothesis.

**Conclusion:** There is extremely strong evidence to suggest that the researcher's claim is *not* valid. The yearly consumption of soft drinks per person is significantly different from 52 gallons. The sample data suggests it is higher.