R=5(c/a-0.3) pleaee help me solve C

Hi! The goal of the Chi-Square Goodness of Fit test is to determine if observed frequencies of a categorical variable align with the expected historical or theoretical values in the population. Having the sales proportions of the top-five compact cars, we compare them against 400 compact car sales data from Chicago to see if there are discrepancies. Specifically, we have: - Chevy Cruze 24% ⟹ P(CC) = 0.24 - Ford Focus 21% ⟹ P(FF) = 0.21 - Hyundai Elantra 20% ⟹ P(HE) = 0.20 - Honda Civic 18% ⟹ P(HC) = 0.18 - Toyota Corolla 17% ⟹ P(TC) = 0.17 The hypotheses established are: H₀: P(CC) = 0.24; P(FF) = 0.21; P(HE) = 0.20; P(HC) = 0.18; P(TC) = 0.17 H₁: There is a discrepancy between expected and observed outcomes. With α set at 0.05, the statistic calculated is based on Oi (observed frequency) and Ei (expected frequency). The initial step involves calculating expected frequencies using: Ei = n * Pi, where Pi is the theoretical proportion for each category stated in the null hypothesis. The test conducted is right-tailed, and so is the p-value, calculated as: P(X²₄ ≥ 11.23) = 1 - P(X²₄ < 11.23) = 1 - 0.98 = 0.02. Since the p-value is lower than α, we reject the null hypothesis, indicating that Chicago's market shares for the five compact cars differ from those reported by Motor Trend.

Response:

404 instances

Detailed breakdown:

Answer:

The value calculated is Z = 2.53, which exceeds 1.96 at a significance level of 0.05

This leads to the rejection of the null hypothesis H₀

Thus, we accept the alternative hypothesis

This implies that individuals participating in Weight Reducers will lose more than 10 pounds

Step-by-step explanation:

Step(i):-

The size of the random sample 'n' = 50

The new weight-loss program, Weight Reducers International, claims that members will shed an average of 10 pounds within the initial two weeks, with a standard deviation of 2.8 pounds.

Population mean 'μ' = 10 pounds

Population standard deviation 'σ' = 2.8 pounds

Sample mean 'x⁻' = 9

Significance level ∝ = 0.05

Step(ii):-

Null hypothesis: H₀: μ < 10

Alternative hypothesis: H₁: μ > 10

Test statistic calculation

The determined Z value is 2.53

In this case, the critical value Z is 1.96 at the 0.05 significance level

Step(iii):-

Calculated Z value of 2.53 is greater than 1.96 at a significance level of 0.05

Consequently, we reject the null hypothesis H₀

We accept the alternative hypothesis

Conclusion:-

We determine that individuals who sign up for Weight Reducers will experience a weight loss exceeding 10 pounds

Answer:

(B) There is a single solution: x = 0.

Step-by-step explanation:

The equation that Kate is attempting to solve is:

Consequently, this equation results in one solution: x = 0.

Answer:

A histogram will be utilized to illustrate the right wrist size of the random sample of newborn infants.

Step-by-step explanation:

A histogram serves as a visual representation of the frequency distribution present within the sample. Given that wrist circumference can take on positive real number values, a histogram with defined class boundaries can be constructed to display the overall distribution of wrist sizes in the graph.

Furthermore, as this distribution is continuous in nature, a histogram proves to be a fitting choice compared to either a bar graph or a stem-and-leaf plot.