Let x be a random variable that represents the pH ofarterial plasma (acidity of
ID: 2913602 • Letter: L
Question
Let x be a random variable that represents the pH ofarterial plasma (acidity of the blood). For healthy adults, themean of the x distrbution is = 7.4(Reference Merck Manual) Anew drug for arthritis has been developed. However, it is thoughtthat this drug may change blood pH. A random sanple of 31 patientswith arthritis took the drug for 3 months. Blood tests showed theaverage pH to be _ ˜8.1 x with sample standard deviation s = 1.9. Use a 5% levelof signifiance to test the claim that the drug has changed(either way) the mean pH level of the blood. Let x be a random variable that represents the pH ofarterial plasma (acidity of the blood). For healthy adults, themean of the x distrbution is = 7.4(Reference Merck Manual) Anew drug for arthritis has been developed. However, it is thoughtthat this drug may change blood pH. A random sanple of 31 patientswith arthritis took the drug for 3 months. Blood tests showed theaverage pH to be _ ˜8.1 x with sample standard deviation s = 1.9. Use a 5% levelof signifiance to test the claim that the drug has changed(either way) the mean pH level of the blood.Explanation / Answer
Let xbe a random variable that represents the pH of arterial plasma(acidity of the blood). For healthy adults, the mean of the xdistribution is = 7.4(Reference Merck Manual) A new drug forarthritis has been developed. However, it is thought that this drugmay change blood pH. A random sample of 31 patients with arthritistook the drug for 3 months. Blood tests showed the average pH to beX_bar ˜ 8.1 with sample standard deviation s = 1.9. Use a 5%level of significance to test the claim that the drug haschanged (either way) the mean pH level of theblood.{Null Hypothesis} = {H0: Mean Blood PH With New Drug EQUALS7.4 }
{Alt Hypothesis} = { Ha: MeanBlood PH With New Drug NOT= 7.4 }
{Test Level Of Significance} = 5%
{Population Mean} = = 7.4
{Sample Size} = N = 31
{Degrees Of Freedom} = df = (31) - 1 = 30
{Sample Mean} = (X_bar) = 8.1
{Sample Std Dev} = s = 1.9
{"t" Test Value} = ( (X_bar)- )/(s/Sqrt[N]) =
= ((8.1)- (7.4))/((1.9)/Sqrt[31])
= 2.051
{P-Value For "t" Test Value(2.051),2-Tail, df=30 (from tables)} = 0.0491
----> {Since testP-Value (0.0491) is LESS than (0.05),
we REJECT Null Hypothesis and acceptAlternate
Hypothesisthat the new drug causes a significant
difference in mean blood PH at the 5%Level Of
Significance.}
.
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