The spreadsheet entitled Compact Cars has sales data for six of the top automobile models, over a six-month period in 2019. There are no sales for the whole/complete of 2019-year sales data meaning what is available is a sample of the 2019-year sales. Your task is to test both: if the mean number of automobile types sold is significantly different for all of the six models AND whether or not the mean number of automobiles sold per month is significantly different among the months (Some of these tests will be taught in a later part of the class under "Hypothesis Testing" & ANOVA). The sample of automobile sales data for 2019 is in the Excel file provided. So, first do the following tests and conclude the findings (these have been taught in class):

Month Chevy Malibu Ford Fusion Hyundai Sonata Honda Accord Toyota Camry VW Passat January 22,161 21,753 21,456 19,791 21,435 20,121 February 18,724 19,835 20,442 21,322 20,235 19,555 March 18,384 17,007 16,163 17,631 17,339 16,456 April 19,837 17,870 15,504 14,950 14,543 14,533 May 17,547 16,597 15,473 16,250 16,177 17,325 June 16,694 16,623 14,745 15,508 15,686 17,343A random sample of nj = 16 communities in western Kansas gave the following rates of hay fever per 1000 population for people under 25 years of age. 121 115 124 99 134 121 110 116 113 96 116 116 135 96 96 116 A random sample of n2 = 14 communities in western Kansas gave the following rates of hay fever per 1000 population for people over 50 years old. 113 86 106 102 113 94 94 108 103 99 78 105 88 100 Assume that the hay fever rate in each age group has an approximately normal distribution. Using the method outlined in Brase and Brase, do the data indicate that the age group over 50 has a lower rate of hay fever? Use a -0,05. Do you reject or fail to reject the null hypothesis? Are the data statistically significant at the a - 0.05 level of significance? Since the p-value is greater than the level of significance, the data are not statistically significant. Thus, we fail to reject the null hypothesis. Since the p-value is less than the level of significance, the data are not statistically significant. Thus, we fail to reject the null hypothesis. Since the p-value is greater than the level of significance, the data are statistically significant. Thus, we fail to reject the null hypothesis. Since the p-value is less than the level of significance, the data are statistically significant. Thus, we reject the null hypothesis. Since the p-value is greater than the level of significance, the data are not statistically significant. Thus, we reject the null hypothesis.2. Suppose you ask a person at many random times to state how happy they are [on some scale}. These scores fora singie person are then averaged to get an average happiness snore [AHIL If you nd the AH score for different people. you'll curiously get different results, but let's suppose that the AH scores for random people are normaily distributed with mean 65 [on some particular happiness instrument}. a. Suppose you are told that the happiest 1% of people score 83 or above. What is the standard deviation of the AH distribution using this happiness instrument? b. A new happiness instrument is designed where AH is normallyI distributed with mean 50 and standard deviation 5. How marryr random peopie would you have to talk to before you met a total ofthree people who scored below 43 on this new instrument? 3. A US company pays interest for $30,000 matured on its bonds held by a German investor. The payment is in Dollars. The German investor sells the Dollars to the European Central Bank (official institution). How do you record this transaction in the US Balance of Payments? Current Account $30,000 Financial Account (Non-official) (a) Financial Account (Official) $30.000 Current Account Financial Account (Non-official) (b) $30.000 $30,000 Financial Account (Official) Current Account $30,000 Financial Account (Non-official) (c) Financial Account (Official) $30,000 Current Account Financial Account (Non-official) (d) $30.000 Financial Account (Official) $30,000 3