(c) Estimate the P-value:- ○ P-value > 0.150- ○ 0.100 < P-value ≤ 0.150- ○ 0.050 < P-value ≤ 0.100- ○ 0.020 < P-value ≤ 0.050- ○ P-value < 0.020Sketch the sampling distribution and show the area corresponding to the P-value.There are four bell-shaped curves displayed, each with a standard normal distribution. The x-axis ranges from -4 to 4 for each graph, and the area under the curve is shaded for the estimate of the P-value.(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?- ○ At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.- ○ At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.- ○ At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.- ○ At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.(e) Interpret your conclusion in the context of the application.- ○ There is sufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood.- ○ There is insufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood. ### Statistical Analysis of Blood pH Levels#### Problem StatementLet \( x \) be a random variable representing the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the \( x \) distribution is \( \mu = 7.4 \). A new drug for arthritis has been developed. It is believed this drug may change blood pH. A random sample of 31 patients with arthritis took the drug for 3 months. Blood tests showed a mean pH of \( \bar{x} = 8.4 \) with a sample standard deviation \( s = 2.5 \). Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood.#### Questions##### (a) Determining the Level of Significance**State the null and alternate hypotheses:**- \( H_0: \mu = 7.4 \); \( H_1: \mu \neq 7.4 \)- \( H_0: \mu = 7.4 \); \( H_1: \mu < 7.4 \)- \( H_0: \mu = 7.4 \); \( H_1: \mu > 7.4 \)- \( H_0: \mu = 7.4 \); \( H_1: \mu \neq 7.4 \)##### (b) Choosing the Sampling Distribution**Which sampling distribution will you use? Explain the rationale for your choice of sampling distribution.**- The Student’s \( t \), since the sample size is large and \( \sigma \) is unknown.- The Student’s \( t \), since the sample size is large and \( \sigma \) is known.- The standard normal, since the sample size is large and \( \sigma \) is known.- The standard normal, since the sample size is large and \( \sigma \) is unknown.**Calculate the sample test statistic:** (Round your answer to three decimal places.)##### (c) Estimating the P-value**Estimate the P-value:**- \( P\text{-value} > 0.150 \)- \( 0.100 < P\text{-value} \leq 0.150 \)- \( 0.050 < P\text

The sample size is n = 31.The sample mean is = 8.4.The sample standard deviation is s = 2.5.We have to test whether the drug has changed the mean pH level of the blood or not.(a). The level of significance is = 0.05.Null Hypothesis:The mean pH level of the blood is 7.4.Alternative Hypothesis:The drug has changed the mean pH level of the blood.Rejection Rule:If p-value (=0.05), then reject the null hypothesis.(b). Sampling distribution:The Student’s t, since the sample size is large and is unknown.It is given that, n = 31, = 8.4, s = 2.5The value of the test statistic is t = 2.227.(c). df = n-1 = 31-1 = 30EXCEL command is “= TDIST (x, deg_freedom, tails)”For p-value, enter in EXCEL as, “= TDIST (2.227, 30, 2)”P-value = 0.0336The p-value estimation is 0.020 < P-value < 0.050.Since p-value (=0.0336) (=0.05), then reject the null hypothesis.(d). Decision:At the = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.(e). Interpretation:There is sufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood.(a). 0.05(b). The Student’s t, since the sample size is large and is unknown.The value of the test statistic is 2.227.(c). 0.020 < P-value < 0.050(d). At the = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.(e). There is sufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood.