Question

listed below are the lead concentrations (n po/g) measured in different ayurveda medicines. ayurveda is a traditional medical system commonly used in india. the lead concentrations listed here are from medicines manufactured in the united states. assume that a simple random sample has been selected. use a 0.01 significance level to test the claim that the mean lead concentration for all such medicines is less than 14.0 po/g. 2.96 0.45 6.03 5.48 20.46 7.48 12.03 20.50 11.46 17.50 identify the null and alternative hypotheses. h₀: μ = 14.0 h₁: μ < 14.0 (type integers or decimals. do not round.) identify the test statistic. (round to two decimal places as needed)

Explanation:

Step1: Calculate the sample mean \(\bar{x}\)

The sample data is \(x = \{2.96,0.45,6.03,5.48,20.46,7.48,12.03,20.50,11.46,17.50\}\).
The formula for the sample mean is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\).
\(\sum_{i=1}^{10}x_i=2.96 + 0.45+6.03+5.48+20.46+7.48+12.03+20.50+11.46+17.50=104.35\)
\(n = 10\), so \(\bar{x}=\frac{104.35}{10}=10.435\)

Step2: Calculate the sample standard deviation \(s\)

The formula for the sample standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
\((x_1-\bar{x})^2=(2.96 - 10.435)^2=(-7.475)^2 = 55.875625\)
\((x_2-\bar{x})^2=(0.45-10.435)^2=(-9.985)^2 = 99.700225\)
\((x_3-\bar{x})^2=(6.03 - 10.435)^2=(-4.405)^2 = 19.404025\)
\((x_4-\bar{x})^2=(5.48-10.435)^2=(-4.955)^2 = 24.552025\)
\((x_5-\bar{x})^2=(20.46-10.435)^2=(10.025)^2 = 100.500625\)
\((x_6-\bar{x})^2=(7.48-10.435)^2=(-2.955)^2 = 8.732025\)
\((x_7-\bar{x})^2=(12.03-10.435)^2=(1.595)^2 = 2.544025\)
\((x_8-\bar{x})^2=(20.50-10.435)^2=(10.065)^2 = 101.303225\)
\((x_9-\bar{x})^2=(11.46-10.435)^2=(1.025)^2 = 1.050625\)
\((x_{10}-\bar{x})^2=(17.50-10.435)^2=(7.065)^2 = 49.914225\)
\(\sum_{i = 1}^{10}(x_i-\bar{x})^2=55.875625+99.700225+19.404025+24.552025+100.500625+8.732025+2.544025+101.303225+1.050625+49.914225 = 463.57665\)
\(s=\sqrt{\frac{463.57665}{9}}\approx7.19\)

Step3: Calculate the test statistic \(t\)

The formula for the \(t\) - test statistic (since the population standard deviation \(\sigma\) is unknown) is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\)
We have \(\bar{x}=10.435\), \(\mu = 14.0\), \(s\approx7.19\), \(n = 10\)
\(t=\frac{10.435 - 14.0}{7.19/\sqrt{10}}=\frac{-3.565}{2.27}\approx - 1.17\)

Answer:

• Null hypothesis: \(H_0:\mu = 14.0\)
• Alternative hypothesis: \(H_1:\mu<14.0\)
• Test statistic: \(-1.17\)