To construct a 90% confidence interval estimate of the average number of store visits based on the provided data, we follow these steps:

Given Data:

Store visits: $6, 10, 4, 5, 7, 4$

Step 1: Compute the sample mean ($\bar{x}$) and sample standard deviation ($s$).

$\bar{x} = \frac{\sum x_i}{n}$

$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Step 2: Determine the critical value ($t^*$) for a 90% confidence level with $n-1 = 5$ degrees of freedom.

Step 3: Calculate the margin of error ($E$).

$E = t^* \times \frac{s}{\sqrt{n}}$

Step 4: Construct the confidence interval.

$\text{Confidence Interval} = \left( \bar{x} - E, \bar{x} + E \right)$

Now, I'll calculate these values step by step.import numpy as np from scipy.stats import t

Given data

store_visits = [6, 10, 4, 5, 7, 4] n = len(store_visits)

Step 1: Calculate sample mean and standard deviation

mean = np.mean(store_visits) std_dev = np.std(store_visits, ddof=1) # Sample standard deviation

Step 2: Determine the critical t-value for 90% confidence level and df = n-1

confidence_level = 0.90 degrees_freedom = n - 1 t_critical = t.ppf((1 + confidence_level) / 2, degrees_freedom)

Step 3: Calculate margin of error

margin_of_error = t_critical * (std_dev / np.sqrt(n))

Step 4: Compute the confidence interval

confidence_interval = (mean - margin_of_error, mean + margin_of_error)

mean, std_dev, t_critical, margin_of_error, confidence_interval(6.0, 2.280350850198276, 2.015048372669157, 1.8759079450505391, (4.124092054949461, 7.875907945050539))(6.0, 2.280350850198276, 2.015048372669157, 1.8759079450505391, (4.124092054949461, 7.875907945050539))success### Results:

1. Sample Mean ($\bar{x}$): $6.0$
2. Sample Standard Deviation ($s$): $2.28$
3. Critical t-value ($t^*$): $2.015$
4. Margin of Error ($E$): $1.876$
5. 90% Confidence Interval: $(4.12, 7.88)$

This means we are 90% confident that the true average number of store visits lies between $4.12$ and $7.88$.

Do you have any questions or want further details? Here are some follow-up questions to consider:

1. How does the sample size affect the confidence interval width?
2. What is the importance of assuming the population is approximately normal?
3. What would change if we were constructing a 95% confidence interval instead?
4. How is the critical t-value determined for different confidence levels?
5. Can this method be applied to datasets with larger or smaller samples?

Tip: Always ensure that the data satisfies the normality assumption when using the t-distribution for small sample sizes.