Math Problem Solver

1.) Given blood pressure measurements before and after drug administration:

Pre-drug: 180,195,164,210,149,194180, 195, 164, 210, 149, 194180,195,164,210,149,194

Post-drug: 110,164,171,180,154,110110, 164, 171, 180, 154, 110110,164,171,180,154,110

Steps:

1. Calculate the differences di=Pre-drug−Post-drugd_i = \text{Pre-drug} - \text{Post-drug}di=Pre-drug−Post-drug.

2. Compute the mean and standard deviation of these differences.

3. Use the t-statistic formula for paired samples: t=dˉsd/nt = \frac{\bar{d}}{s_d / \sqrt{n}}t=sd/ndˉ where dˉ\bar{d}dˉ is the mean difference, sds_dsd is the standard deviation of differences, and nnn is the sample size.

4. Compare the calculated ttt value with the critical ttt-value for n−1n - 1n−1 degrees of freedom.

2.) The sales manager assumes a normal distribution with:

• Mean μ=2,500,000\mu = 2,500,000μ=2,500,000

• Standard deviation σ=250,000\sigma = 250,000σ=250,000

i. Probability Sales Exceed 3,050,000

Use the z-score formula:

z=X−μσz = \frac{X - \mu}{\sigma}z=σX−μ

Substitute X=3,050,000X = 3,050,000X=3,050,000 and find the probability that sales exceed this amount.

ii. Probability Sales Fall Between 2,000,000 and 3,000,000

Calculate the z-scores for X=2,000,000X = 2,000,000X=2,000,000 and X=3,000,000X = 3,000,000X=3,000,000, then find the probability that sales are within this range by finding the area under the normal curve between these two z-scores.

iii. Sales Level with 12% Chance of Being Exceeded

Find the sales value XXX such that the probability of exceeding XXX is 0.12. Using the z-table, locate the z-score corresponding to the 88th percentile and solve for XXX.

3.) The data provided is:

• Females: Daily = 24, Weekly = 24

• Males: Daily = 10, Weekly = 35

i. Hypotheses

• Null hypothesis H0H_0H0: There is no association between gender and shopping habits.

• Alternative hypothesis H1H_1H1: There is an association between gender and shopping habits.

ii. Chi-Square Test

1. Construct the contingency table.

2. Calculate the expected frequencies for each cell.

3. Compute the chi-square statistic: χ2=∑(O−E)2E\chi^2 = \sum \frac{(O - E)^2}{E}χ2=∑E(O−E)2 where OOO is the observed frequency and EEE is the expected frequency.

4. Compare the calculated chi-square value with the critical value at α=0.01\alpha = 0.01α=0.01.

4.) Before and after advertising data:

• Pre-advertising mean = 140 dozen

• Post-advertising sample mean = 148 dozen, n=25n = 25n=25, s=15s = 15s=15

Use a one-sample t-test to check if the post-campaign sales differ significantly from the pre-campaign mean.

Steps:

1. Calculate the t-statistic: t=Xˉ−μs/nt = \frac{\bar{X} - \mu}{s / \sqrt{n}}t=s/nXˉ−μ where Xˉ=148\bar{X} = 148Xˉ=148, μ=140\mu = 140μ=140, s=15s = 15s=15, and n=25n = 25n=25.

2. Compare with the critical ttt-value for n−1n - 1n−1 degrees of freedom.

5.) Data:

• Before tax: p1=16500p_1 = \frac{16}{500}p1=50016

• After tax: p2=3100p_2 = \frac{3}{100}p2=1003

To determine if there’s a significant decrease:

1. Calculate the pooled proportion: p=x1+x2n1+n2p = \frac{x_1 + x_2}{n_1 + n_2}p=n1+n2x1+x2

2. Compute the z-score: z=p1−p2p(1−p)(1n1+1n2)z = \frac{p_1 - p_2}{\sqrt{p(1 - p)(\frac{1}{n_1} + \frac{1}{n_2})}}z=p(1−p)(n11+n21)p1−p2

3. Compare with the critical z-value at the desired significance level.

6.) Data for blood sugar levels across four antidotes (A1, A2, A3, A4):

• Compute the ANOVA to compare the mean effectiveness of each antidote.

Steps:

1. Calculate the total sum of squares (SST), the sum of squares between groups (SSB), and within groups (SSW).

2. Determine the F-ratio: F=MSBMSWF = \frac{\text{MSB}}{\text{MSW}}F=MSWMSB

3. Compare the F-value with the critical F-value at the given degrees of freedom.

Each analysis can be carried out step-by-step, or let me know if you'd like to see the calculations for a specific question!

re write the calculations