The null and alternative hypotheses are given. Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. What parameter is being​ tested?   H0​: σ=4.2 H1​: σ≠4.2

Use technology to find the​ P-value for the hypothesis test described below.   The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean is μ=16.00 Mbps. The sample size is n=19 and the test statistic is t=1.224.

review(31a) MAKE SURE YOU SELECT TWO ANSWERS FOR THIS QUESTION.. Type of test and hypotheses. Professor dill  wonders if male and female earn significantlly differnent grades, when taking the same course with the same instructor. He randomly samples 160 males. and 100 females. The males receive 20 A's, 32 B's 70 C's  30 D's, anf 8 F's.The female receive 20'A's, 28 B's, 30 C's, and 10 D's. and 12 F's; His hypotheses is that gender and grades are related. Choose the type of test you should choose to test the hypotheses in symbol notation( or in words if symbol is not appropriate) (a) Z test (b)single sample t test (c)dependent t test (d) independent t test (e) Pearson correlation coeffieient. (f) ANOVA (g) chi Square goodness  of fit test (h)Chi Square test of independence  (i) Ho: u1-u2=0     H1: u1-u2!=0 (j) Ho: u1-u2 =0     H1:u1-u2<0 (k) Ho: u1-u2 >= 0      H1: u1=u2<0 (l) Ho : Rho=0     H1: Rho!=0 (m) Ho: Rho >= 0       H1: Rho <0 (n) Ho:Rho <= 0      H1: Rho >…

Read through this scenario and look at the data that was collected.  State the null and all possible research hypotheses.  Review the results below (I used SPSS) and answer the questions that follow. Scenario:  A researcher wants to see if gender and / or income affects the total amount of help given to a stranger who is sitting on the side of a busy road with a sign asking for help.  The independent variables are gender, income, and the interaction of gender and income.  The dependent variable is total help. He wants to know if one or both factors – or the interaction of the two - affects the total amount of help offered.  Because he is analyzing two independent variables (gender and income), he used a factorial ANOVA.  His results show the main effect of each of the independent variables on the dependent variable (total help) and the interaction effect.  The researcher is using a 95% confidence interval which means that he wants to be at least 95% sure that his independent variables…

Consider the following data: - 4 - 3 - 1 P(X = x) 0.2 0.1 0.2 0.1 0.4 Step 4 of 5: Find the value of P(X > -2). Round your answer to one decimal place.

Use technology to find the​ P-value for the hypothesis test described below. The claim is that for a smartphone​ carrier's data speeds at​ airports, the mean is μ=14.00Mbps. The sample size is n=25and the test statistic is t=−1.313.

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Question list O O Question 21 O Question 22 O Question 23 Question 24 Question 25 Question 26 uestion 27 estion 28 stion 29 ↓ K The P-value for a hypothesis test is shown. Use the P-value to decide whether to reject Ho when the level of significance is (a) a = 0.01, (b) a = 0.05, and (c) a=0.10. P=0.0138 (a) Do you reject or fail to reject Ho at the 0.01 level of significance? OA. Fail to reject Ho because the P-value, 0.0138, is less than α=0.01. OB. Fail to reject Ho because the P-value, 0.0138, is greater than a = 0.01. OC. Reject Ho because the P-value, 0.0138, is less than a = 0.01. OD. Reject Ho because the P-value, 0.0138, is greater than a = 0.01. (b) Do you reject or fail to reject Ho at the 0.05 level of significance? O A. Reject Ho because the P-value, 0.0138, is less than a = 0.05. OB. Reject Ho because the P-value, 0.0138, is greater than a = 0.05. OC. Fail to reject Ho because the P-value, 0.0138, is less than a = 0.05. O D. Fail to reject Ho because the P-value, 0.0138,…

Please finish the R code to conduct the hypothesis testing. I need the code for the p-value for the F test, tstat, and the p value for tstat. Write the conclusion to the hypothesis testing once completed.  # Example 11.1# simple linear regression# H0: B=0 (there is no linear correlation)# H1: B≠0 (there is a linear correlation)data <- data.frame(estriol=c(7,9,9,12,14,16,16,14,16,16,17,19,21,24,15,16,17,25,27,15,15,15,16,19,18,17,18,20,22,25,24),birthweight=c(25,25,25,27,27,27,24,30,30,31,30,31,30,28,32,32,32,32,34,34,34,35,35,34,35,36,37,38,40,39,43))linear <- lm(birthweight~estriol, data = data)summary(linear)# This gives the results for a and b and the F statistic. # Call:#   lm(formula = birthweight ~ estriol, data = data)# # Residuals:#   Min      1Q  Median      3Q     Max # -8.1200 -2.0381 -0.0381  3.3537  6.8800 # # Coefficients:#   Estimate Std. Error t value Pr(>|t|)    # (Intercept)  21.5234     2.6204   8.214 4.68e-09 ***#   estriol       0.6082     0.1468   4.143…