Are Florida Lakes Acidic or Alkaline?The pH of a liquid is a measure of its acidity or alkalinity. Pure water has a pH of 7, which is neutral. Solutions with a pH less than 7 areacidic while solutions with a pH greater than 7 are basic, or alkaline. The dataset Florida Lakes gives information, including pH values,for a sample of lakes in Florida. Computer output of descriptive statistics for the pH variable is shown:Descriptive Statistics: pHVariable N N*pH53 0n=x=SEMean6.591 0.177Click here for the dataset associated with this question.Enter the exact answers.S=MeanO Ee Textbook and MediaStDev Minimum QI Median1.288(a) How many lakes are included in the dataset? What is the mean pH value? What is the standard deviation?BE35.8003.6006.800CF903Maximum7:450 9.100F10F13F12 (b) Use the descriptive statistics above to conduct a hypothesis test to determine whether there is evidence that average pH inFlorida lakes is different from the neutral value of 7.State the null and alternative hypotheses.O Ho: H = 7 vs HA: μ # 7O Ho: H =O Ho: H =O Ho: H =O Ho: H =6.591 vs HA:μ # 6.5917 vs H₁: H > 76.591 vs HA: μ<6.5917 VS HA: H <7

a)The descriptive statistics for the pH values are given.From the output, the number of lakes that are included in the dataset is 53 (N), the mean pH value is 6.591 (Mean) and the standard deviation is 1.288 (StDev).That is,n = 53x̄ = 6.591s = 1.288.b)Suppose μ is the population average pH in the lakes.It is needed to test whether the average pH in lakes is different from the neutral value of 7.The null and alternative hypotheses are,H0: μ = 7 vs HA: μ ≠ 7The test is both tailed.In this case, the sample size is large (> 30) and the population standard deviation is not mentioned. Samples are drawn randomly and independently.Hence, one sample t-test for a population mean can be used.The degrees of freedom is, df = n – 1 = 53 – 1 = 52.The test statistic is, The test statistic is –2.312.Suppose level of significance is α = 0.05.Hence, the p-value using the EXCEL formula =T.DIST.2T(2.312,52), is approximately 0.0248.The excel output is, Thus, the p-value is 0.0248.Decision rule:If p-value is less than or equal to the level of significance, reject the null hypothesis.Otherwise fail to reject the null hypothesis.Conclusion:The p-value of 0.0248 is less than the level of significance of 0.05.Hence, reject the null hypothesis.There is sufficient evidence to conclude that the average pH in lakes is different from the neutral value of 7, at 5% level of significance.n = 53x̄ = 6.591s = 1.288The null and alternative hypotheses are,H0: μ = 7 vs HA: μ ≠ 7There is sufficient evidence to conclude that the average pH in lakes is different from the neutral value of 7, at 5% level of significance.