Steps to Solve:

We are tasked to compute:

1. Sample proportion ($\hat{p}$).
2. Test statistic ($z$).
3. p-value for the hypothesis test.

The hypotheses are:

• Null Hypothesis ($H_0$): $p \geq 0.8$
• Alternative Hypothesis ($H_a$): $p < 0.8$

We use a significance level $\alpha = 0.05$.

Step 1: Identify packages with 3 ounces or more.

We count how many packages from the given weights are $\geq 3.0$ ounces.

Step 2: Compute the sample proportion ($\hat{p}$):

$\hat{p} = \frac{\text{Number of packages } \geq 3.0}{\text{Total number of packages}}$

Step 3: Compute the test statistic ($z$):

The formula is: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$ Where:

• $p_0 = 0.8$ (claimed proportion),
• $n = 50$ (sample size).

Step 4: Compute the p-value:

The p-value is the probability of observing a value as extreme as $\hat{p}$, assuming the null hypothesis is true. It is calculated as: $p\text{-value} = P(Z < z)$

Let's compute this using the provided data. numpy as np from scipy.stats import norm

Given data (weights of chocolate stars in packages)

weights = [ 2.872136, 4.24198, 3.117765, 3.680094, 3.580635, 2.991352, 3.204946, 2.24881, 3.468928, 3.592362, 3.065978, 2.16143, 3.789306, 4.181118, 3.280708, 3.340212, 3.563963, 3.708506, 4.246951, 3.492321, 3.111458, 2.97365, 3.589066, 3.350737, 2.609161, 2.61035, 3.946645, 2.791369, 3.237388, 3.440239, 2.533146, 3.803544, 3.052931, 2.960728, 3.848325, 3.731875, 3.017454, 3.787895, 3.221476, 2.953563, 2.490938, 3.774448, 3.05537, 3.678658, 2.183067, 3.08827, 2.834771, 3.573603, 3.282198, 3.623261 ]

Parameters

p0 = 0.8 # Null hypothesis proportion n = 50 # Sample size

Step 1: Calculate number of packages with weight >= 3.0

count_gte_3 = sum(weight >= 3.0 for weight in weights)

Step 2: Calculate sample proportion (p-hat)

p_hat = count_gte_3 / n

Step 3: Calculate test statistic (z)

z = (p_hat - p0) / np.sqrt((p0 * (1 - p0)) / n)

Step 4: Calculate p-value (left-tailed test)

p_value = norm.cdf(z)

p_hat, z, p_value### Results:

1. Sample proportion ($\hat{p}$): $0.720$
2. Test statistic ($z$): $-1.414$
3. p-value: $0.079$

Conclusion:

• Since the p-value ($0.079$) is greater than the significance level $\alpha = 0.05$, we fail to reject the null hypothesis.
• Interpretation: There is not enough evidence to conclude that the proportion of packages with 3 ounces or more of chocolate stars is less than 0.8.

Follow-up Questions:

1. How would the conclusion change if the significance level were $0.10$ instead of $0.05$?
2. What is the critical value of $z$ at $\alpha = 0.05$ for a one-tailed test?
3. How does increasing the sample size $n$ impact the results of the test?
4. If the true proportion $p$ were $0.75$, what is the power of this test?
5. How would a two-tailed test alter the hypothesis and the interpretation of results?

Tip:

When interpreting hypothesis tests, always check the p-value against the significance level and confirm that the assumptions (e.g., normality of sampling distribution) are met.